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SCHOOL SECTION 49

o Introduction : :

An equation involving one variable with highest power of variable as 2 is

called a quadratic equation.

Example : x2 + 6x + 7 = 0, 2x2 + 7 = 0

The general form of quadratic equation is ax2 + bx + c = 0 where a, b and c

are real numbers and a 0.

Things To Remember :

• Highest power of variable has to be 2.

• Equation has to be in simplified form.

• Simplification implies having all the terms involving variable in the

numerator. If variables are in the denominator, then each term of equation

has to be multiplied by the term of denominator so that variable in the

denominator gets cancelled.

EXERCISE - 2.1 (TEXT BOOK PAGE NO. 36) :1. Which of the following are quadratic equations ?(i) 11 = – 4x2 – x3 (1 mark)Sol. 11 = 4x2 – x3

x3 + 4x2 + 11 = 0Here maximum index of the variable x is 3.So it is not a quadratic equation.

(ii)3

–4

y2 = 2y + 7 (1 mark)

Sol.3

–4

y2 = 2y + 7

3

–4

y2 – 2y – 7 = 0

Here a = 3

–4

, b = –2, c = –7 are real numbers

where a 0So it is a quadratic equation in variable y.

(iii) (y – 2) (y + 2) = 0 (1 mark)Sol. (y – 2) (y + 2) = 0

(y)2 – (2)2 = 0 [ a2 – b2 = (a + b) (a – b)] y2 – 4 = 0 y2 + 0y – 4 = 0

Here a = 1, b = 0, c = – 4 are real numberswhere a 0So it is a quadratic equation in variable y.

Quadratic Equations2.

ALGEBRA MT EDUCARE LTD.

SCHOOL SECTION50

(iv)3y – 4 = y (1 mark)

Sol.3

y – 4 = y

Multiplying throughout by y, we get,3 – 4y = y2

– y2 – 4y + 3 = 0Here a = –1, b = – 4, c = 3 are real numberswhere a 0So it is a quadratic equation in variable y.

(v) m3 + m + 2 = 4m (1 mark)Sol. m3 + m + 2 = 4m

m3 + m – 4m + 2 = 0 m3 – 3m + 2 = 0

Here the maximum index of the variable m is 3.So it is not a quadratic equation.

(vi) n – 3 = 4n (1 mark)Sol. n – 3 = 4n

n – 4n – 3 = 0 – 3n – 3 = 0 0n2 – 3n – 3 = 0

Here a = 0So it is not a quadratic equation.

(vii) y2 – 4 = 11y (1 mark)Sol. y2 – 4 = 11y

y2 – 11y – 4 = 0Here a = 1, b = –11, c = – 4 are real numberswhere a 0So it is a quadratic equation in variable y.

(viii) z – 7z

= 4z + 5 (1 mark)

Sol. z – 7

z = 4z + 5

Multiplying throughout by z, we get,z2 – 7 = 4z2 + 5z

z2 – 4z2 – 5z – 7 = 0 –3z2 – 5z – 7 = 0

Here a = –3, b = –5, c = –7 are real numberswhere a 0So it is a quadratic equation in variable z.

(ix) 3y2 – 7 = 3 y (1 mark)

Sol. 3y2 – 7 = 3 y

3y2 – 3 y – 7 = 0

Here a = 3, b = – 3 , c = –7 are real numbers

where a 0So it is a quadratic equation in variable y.

MT EDUCARE LTD. ALGEBRA

SCHOOL SECTION 51

(x)2

2

q – 4q = –3 (1 mark)

Sol.q – 4

q

2

2 = –3

q2 – 4 = –3q2

q2 + 3q2 – 4 = 0 4q2 – 4 = 0 4q2 + 0q – 4 = 0

Here a = 4, b = 0, c = – 4 are real numberswhere a 0So it is a quadratic equation in variable q.

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :4. Which of the following equations are quadratic ?(i) 13 = – 5y2 – y3 (1 mark)Sol. 13 = – 5y2 – y3

y3 + 5y2 + 13 = 0Here maximum index of the variable y is 3.So it is not a quadratic equation.

(ii) –53

x2 = 2x + 9 (1 mark)

Sol. –5

3x2 = 2x + 9

–5

3 x2 – 2x – 9 = 0

Here a = –5

3, b = – 2, c = – 9 are real numbers.

Where a 0.So it is a quadratic equation in variable x.

(iii) (x + 3) (x – 4) = 0 (1 mark)Sol. (x + 3) (x – 4) = 0

x (x – 4) + 3 (x – 4) = 0 x2 – 4x + 3x – 12 = 0 x2 – x – 12 = 0

Here a = 1, b = – 1, c = – 12 are real numbers.Where a 0.So it is a quadratic equation in variable x.

(iv)5x

– 3 = x2 (1 mark)

Sol.5

x – 3 = x2

Multiplying throughout by x, we get;5 – 3x = x3

0 = x3 + 3x – 5 x3 + 3x – 5 = 0

Here maximum index of the variable x is 3.So it is not a quadratic equation.


SCHOOL SECTION52

(v) n3 – n + 4 = n3 (1 mark)Sol. n3 – n + 4 = n3

n3 – n3 – n + 4 = 0 – n + 4 = 0

Here a = 0So it is not a quadratic equation.

(vi) n – 3 = 4n2 (1 mark)Sol. n – 3 = 4n2

– 4n2 + n – 3 = 0Here a = – 4, b = 1, c = – 3 are real numbers.Where a 0So it is a quadratic equation in variable n.

(vii) x2 + 4x = 11 (1 mark)Sol. x2 + 4x = 11

x3 + 4x – 11 = 0Here a = 1, b = 4, c = – 11 are real numbers.Where a 0So it is a quadratic equation in variable x.

(viii) m – 5m

= 4m + 5 (1 mark)

Sol. m – 5

m = 4m + 5

Multiplying throughout by m, we get;m2 – 5 = 4m2 + 5m

0 = 4m2 – m2 + 5m + 5 3m2 + 5m + 5 = 0

Here a = 3, b = 5, c = 5 are real numbers.Where a 0So it is a quadratic equation in variable m.

o Standard form of a quadratic equations :The standard form of a quadratic equation in variable x is ax2 + bx + c = 0where a, b and c are real numbers and a 0.

EXERCISE - 2.1 (TEXT BOOK PAGE NO. 36) :2. Write the following quadratic equations in standard form

ax2 + bx + c = 0(i) 7 – 4x –x2 = 0 (1 mark)Sol. 7 – 4x – x2 = 0

– x2 – 4x + 7 = 0Multiplying throughout by –1, we get,

x2 + 4x – 7 = 0

(ii) 3y2 = 10y + 7 (1 mark)Sol. 3y2 = 10y + 7

3y2 – 10y – 7 = 0

(iii) (m + 4) (m – 10) = 0 (1 mark)Sol. (m + 4) (m – 10) = 0

m(m – 10) + 4(m – 10) = 0 m2 – 10m + 4m – 40 = 0 m2 – 6m – 40 = 0


SCHOOL SECTION 53

(iv) p(p – 6) = 0 (1 mark)Sol. p(p – 6) = 0

p2 – 6p = 0 p2 – 6p + 0 = 0

(v)2x

25 – 4 = 0 (1 mark)

Sol.x

25

2

– 4 = 0

Multiplying throughout by 25, we get,x2 – 100 = 0

x2 + 0x – 100 = 0

(vi) n – 7n

= 4 (1 mark)

Sol. n – 7

n = 4

Multiplying throughout by n, we get,n2 – 7 = 4n

n2 – 4n – 7 = 0

(vii) y2 – 9 = 13y (1 mark)Sol. y2 – 9 = 13y

y2 – 13y – 9 = 0

(viii) 2z – 5z

= z – 6 (1 mark)

Sol. 2z – 5

z = z – 6

Multiplying throughout by z, we get, 2z2 – 5 = z2 – 6z

2z2 – z2 + 6z – 5 = 0 z2 + 6z – 5 = 0

(ix) x2 = –7 – 10 x (1 mark)

Sol. x2 = –7 – 10 x

x2 + 10 x + 7 = 0

(x)2

2

m +5m

= –3 (1 mark)

Sol.m 5

m

2

2 = –3

m2 + 5 = –3m2

m2 + 3m2 + 5 = 0 4m2 + 5 = 0 4m2 + 0m+ 5 = 0

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :6. Write the quadratic equations in standard from ax2 + bx + c = 0(i) 8 – 3x – 4x2 = 0 (1 mark)Sol. 8 – 3x – 4x2 = 0

0 = 4x2 + 3x – 8 4x2 + 3x – 8 = 0


SCHOOL SECTION54

(ii) 3y2 = 8y – 5 (1 mark)Sol. 3y2 = 8y – 5

3y2 – 8y + 5 = 0

(iii) (x + 5) (x – 11) = 0 (1 mark)Sol. (x + 5) (x – 11) = 0

x2 – 11x + 5x – 55 = 0 x2 – 6x – 55 = 0

(iv) m (m – 7) = 0 (1 mark)Sol. m (m – 7) = 0

m2 – 7m = 0 m2 – 7m + 0 = 0

(v)2y

23 – 3 = 0 (1 mark)

Sol.2y

23 – 3 = 0

Multiplying throughout by 23, we gety2 – 69 = 0

y2 + 0y – 69 = 0

(vi) x – 6x

= 5 (1 mark)

Sol. x – 6

x = 5

Multiplying throughout by x, we getx2 – 6 = 5x

x2 – 5x – 6 = 0

(vii) – x2 – 5 = 16x (1 mark)Sol. – x2 – 5 = 16x

0 = x2 + 16x + 5 x2 + 16x + 5 = 0

(viii) y – 5y = 3y – 7 (1 mark)

Sol. y – 5

y = 3y – 7

Multiplying throughout by y, we gety2 – 5 = 3y2 – 7y

0 = 3y2 – y2 – 7y + 5 0 = 2y2 – 7y + 5 2y2 – 7y + 5 = 0

(ix) y2 = 5 – 7 y (1 mark)

Sol. y2 = 5 – 7 y

y2 + 7 y – 5 = 0


SCHOOL SECTION 55

(x)2x – 7x

= 7 (1 mark)

Sol.x – 7

x

2

= 7

x2 – 7 = 7x x2 – 7x – 7 = 0

(xi) 4z2 + 5 = 3 z (1 mark)

Sol. 4z2 + 5 = 3 z

4z2 – 3 z + 5 = 0

(xii)2

2

p 53p

+ = – 7 (1 mark)

Sol.p + 5

3p

2

2 = – 7

p2 + 5 = – 7 × 3p2

p2 + 5 = – 21p2

p2 + 21p2 + 5 = 0 22p2 + 0p + 5 = 0

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :1. Find the values of a, b, c for following quadratic equations by comparing

with standard form :(i) x2 + 2x + 1 = 0 (1 mark)Sol. x2 + 2x + 1 = 0

Comparing with ax2 + bx + c = 0a = 1, b = 2, c = 1

(ii) 2x2 – x + 3 = 0 (1 mark)Sol. 2x2 – x + 3 = 0

Comparing with ax2 + bx + c = 0a = 2, b = – 1 c = 3

(iii) x2 – x – 3 = 0 (1 mark)Sol. x2 – x – 3 = 0

Comparing with ax2 + bx + c = 0a = 1, b = – 1, c = – 3

(iv) x2 + 5x – 4 = 0 (1 mark)Sol. x2 + 5x – 4 = 0

Comparing with ax2 + bx + c = 0a = 1, b = 5, c = – 4

(v) x2 – 7x + 4 = 0 (1 mark)Sol. x2 – 7x + 4 = 0

Comparing with ax2 + bx + c = 0a = 1, b = – 7, c = 4


SCHOOL SECTION56

(vi) 4x2 – 9x + 1 = 0 (1 mark)Sol. 4x2 – 9x + 1 = 0

Comparing with ax2 + bx + c = 0a = 4, b = – 9, c = 1

o Roots/solution of a quadratic equation :Any value of the variable which when substituted in a given quadraticequation makes the left hand side of the equation equal to its right handside, is said to be a solution of the equation.A solution of a quadratic equation is also called a root of that equation.

The set of roots of a quadratic equation is called the solution set of thequadratic equation.For example : Consider x2 – 4x + 3 = 0 take sum real numbers as x = 1, 2and 3 and see whether the equation is satisfied or not.(i) L.H.S. = 12 – 4 (1) + 3 = 1 – 4 + 3 = –3 + 3 = 0 L.H.S. = R.H.S. x = 1 is the root of the given quardratic equation(ii) L.H.S. = (2)² – 4 (2) + 3 = 4 – 8 + 3 = – 4 + 3 = –1 L.H.S. R.H.S. x = 2 is not the root of the given quardratic equation(iii) L.H.S. = (3)² – 4 (3) + 3 = 9 – 12 + 3 = 12 – 12 = 0 L.H.S. = R.H.S. x = 3 is the root of the given quardratic equation

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :

3. Determine whether the given values of ‘x’ is a roots of given quadratic equation.(i) x2 – 2x + 1 = 0, x = 1 (1 mark)Sol. x2 – 2x + 1 = 0, x = 1

Putting x = 1 in L.H.S. we get,L.H.S. = (1)2 – 2 (1) + 1

= 1 – 2 (1) + 1= 2 – 2= 0= R.H.S.

L.H.S. = R.H.S.Thus equation is satisfied.So 1 is the root of the given quadratic equation.

(ii) x2 + 2x + 1 = 0, x = – 1 (1 mark)Sol. x2 + 2x + 1 = 0, x = – 1

Putting x = – 1 in L.H.S. we get,L.H.S. = (– 1)2 + 2 (– 1) + 1

= 1 – 2 + 1= 2 – 2= 0= R.H.S.

L.H.S. = R.H.S.Thus equation is satisfied.So – 1 is the root of the given quadratic equation.


SCHOOL SECTION 57

(iii) x2 – x = 0, x = 0 (1 mark)Sol. x2 – x = 0, x = 0

Putting x = 0 in L.H.S., we get,L.H.S. = (0)2 – 0

= 0 – 0= 0= R.H.S.


(iv) x2 + x – 1 = 0, x = 2 (1 mark)Sol. x2 + x – 1 = 0, x = 2

Putting x = 2 in L.H.S., we get,L.H.S. = (2)2 + 2 – 1

= 4 + 1= 5 R.H.S.

L.H.S. R.H.S.Thus equation is not satisfied.So 2 is not the root of the given quadratic equation.

(v) x2 – 4x + 4 = 0, x = 0 (1 mark)Sol. x2 – 4x + 4 = 0, x = 0

Putting x = 0 in L.H.S., we get,L.H.S. = (0)2 + 4 (0) + 4

= 0 – 0 + 4= 4 R.H.S.


(vi) x2 – 4x + 1 = 0, x = 1 (1 mark)Sol. x2 – 4x + 1 = 0, x = 1

Putting x = 1 in L.H.S., we get,L.H.S. = (1)2 – 4 (1) + 1

= 1 – 4 + 1= 2 – 4= – 2 R.H.S.


EXERCISE - 2.2 (TEXT BOOK PAGE NO. 38) :1. In each of the examples given below determine whether the values

given against each of the quadratic equation are the roots of the equationor not.

(i) x2 + 3x – 4 = 0, x = 1, –2, – 3 (3 marks)Sol. a) By putting x = 1 in L.H.S. we get

L.H.S. = (1)2 + 3(1) – 4= 1 + 3 – 4= 4 – 4= 0


SCHOOL SECTION58

= R.H.S. L.H.S. = R.H.S.

Thus equation is satisfied.So 1 is the root of the given quadratic equation.

b) By putting x = –2 in L.H.S. we getL.H.S. = (–2)2 + 3(–2) – 4

= 4 – 6 – 4= – 6 R.H.S.

L.H.S. R.H.S.Thus equation is not satisfied.So –2 is not the root of the given quadratic equation.

c) By putting x = –3 in L.H.S. we getL.H.S. = (–3)2 + 3(–3) – 4

= 9 – 9 – 4= – 4 R.H.S.


(ii) 4m2 – 9 = 0, m = 2, 23

, 32

(3 marks)

Sol. a) By putting m = 2 in L.H.S. we getL.H.S. = 4(2)2 – 9

= 4(4) – 9= 16 – 9= 7 R.H.S.


b) By putting m = 2

3 in L.H.S. we get

L.H.S. = 42

3

2

– 9

= 4 × 4

9 – 9

=16

9 – 9

=16–81

9

=–65

9 R.H.S.

L.H.S. R.H.S.Thus equation is not satisfied.

So 2

3 is not the root of the given quadratic equation.


SCHOOL SECTION 59

c) By putting m = 3

2 in L.H.S. we get

L.H.S. = 43

2

2

– 9

= 4 × 9

4 – 9

= 9 – 9= 0= R.H.S.

L.H.S. = R.H.S.Thus equation is satisfied.

So 3

2 is the root of the given quadratic equation.

(iii) x2 + 5x – 14 = 0, x = 2 , –7, 3 (3 marks)

Sol. a) By putting x = 2 in L.H.S. we get

L.H.S. = ( 2 )2 + 5( 2 ) – 14

= 2 + 5 2 – 14

= 5 2 – 12 R.H.S.


b) By putting x = –7 in L.H.S. we getL.H.S. = (–7)2 + 5(–7) – 14

= 49 – 35 – 14= 49 – 49= 0= R.H.S.

L.H.S. = R.H.S.Thus equation is satisfied.So –7 is the root of the given quadratic equation.

c) By putting x = 3 in L.H.S. we getL.H.S. = (3)2 + 5(3) – 14

= 9 + 15 – 14= 24 – 14= 10 R.H.S.


(iv) 2p2 + 5p – 3 = 0, p = 1, 12

, –3 (3 marks)

Sol. a) By putting p = 1 in L.H.S. we getL.H.S. = 2(1)2 + 5(1) – 3

= 2(1) + 5 – 3= 2 + 5 – 3= 7 – 3= 4 R.H.S.


SCHOOL SECTION60


b) By putting p = 1

2 in L.H.S. we get

L.H.S. = 21

2

2

+ 51

2 – 3

= 2 × 1

4 + 5 ×

1

2 – 3

=1

2 +

5

2 – 3

=6

2 – 3

= 3 – 3= 0= R.H.S.


So 1


c) By putting p = – 3 in L.H.S. we getL.H.S. = 2(– 3)2 + 5(– 3) – 3

= 2(9) – 15 – 3= 18 – 18= 0= R.H.S.


(v) n2 + 4n = 0, n = 0, – 2, – 4 (3 marks)Sol. a) By putting n = 0 in L.H.S. we get

L.H.S. = (0)2 + 4(0)= 0 + 0= 0= R.H.S.


b) By putting n = –2 in L.H.S. we getL.H.S. = (– 2)2 + 4(– 2)

= 4 – 8= – 4 R.H.S.



SCHOOL SECTION 61

c) By putting n = –4 in L.H.S. we getL.H.S. = (– 4)2 + 4(– 4)

= 16 – 16= 0= R.H.S.


PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :10. In each of the examples given below determine whether the values

given against each of the quadratic equation are the roots of the equationor not.

(i) y2 + 5y + 6 = 0; y = 4, – 2, – 3 (3 marks)Sol. y2 + 5y + 6 = 0; y = 4, – 2, – 3

(a) By putting y = 4 in L.H.S. we getL.H.S. = (4)2 + 5 (4) + 6

= 16 + 20 + 6= 42 R.H.S.


(b) By putting y = – 2 in L.H.S. we getL.H.S. = (– 2)2 + 5 (– 2) + 6

= 4 – 10 + 6= 10 – 10= 0= R.H.S.

L.H.S. R.H.S.Thus equation is satisfied.So – 2 is the root of the given quadratic equation.

(c) By putting y = – 3 in L.H.S. we getL.H.S. = (– 3)2 + 5 (– 3) + 6

= 9 – 15 + 6= 15 – 15= 0= R.H.S.


(ii) 25x2 – 36 = 0; x = 2, 65

, – 65

(3 marks)

Sol. 25x2 – 36 = 0; x = 2, 6

5,

– 6

5(a) By putting x = 2 in L.H.S. we get

L.H.S. = 25(2)2 – 36= 25 (4) – 36= 100 – 36= 64


SCHOOL SECTION62

R.H.S. L.H.S. R.H.S.

Thus equation is not satisfied.So 2 is not the root of the given quadratic equation.

(b) By putting x = 6

5 in L.H.S. we get

L.H.S. = 25 6

5

2

– 36

= 25 × 36

25 – 36

= 36 – 36= 0= R.H.S.

L.H.S. R.H.S.Thus equation is satisfied.

So 6


(c) By putting x = 6

5

in L.H.S. we get

L.H.S. = 25 6

5

2

– 36

= 25 × 36

25 – 36

= 36 – 36= 0= R.H.S.

L.H.S. R.H.S.Thus equation is satisfied.

So 6

5

is the root of the given quadratic equation.

(iii) x2 – 7x – 18 = 0; x = 3 , – 2, 4 (3 marks)

Sol. x2 – 7x – 18 = 0; x = 3 , – 2, 4

(a) By putting x = 3 in L.H.S. we get

L.H.S. = ( 3 )2 – 7 ( 3 ) – 18

= 3 – 7 3 – 18

= – 15 – 7 3

R.H.S. L.H.S. R.H.S.

Thus equation is not satisfied.

So 3 is not the root of the given quadratic equation.


SCHOOL SECTION 63

(b) By putting x = – 2 in L.H.S. we getL.H.S. = (– 2)2 – 7 (– 2) – 18

= 4 + 14 – 18= 18 – 18= 0= R.H.S.


(c) By putting x = 4 in L.H.S. we getL.H.S. = (4)2 – 7 (4) – 18

= 16 – 28 – 18= 16 – 46= – 30 R.H.S.


(iv) 3x2 – 8x – 3 = 0; x = 9,12

, 3 (3 marks)

Sol. 3x2 – 8x – 3 = 0;x = 9,1

2, 3

(a) By putting x = 9 in L.H.S. we getL.H.S. = 3 (9)2 – 8 (9) – 3

= 3 (81) – 72 – 3= 243 – 75= 168 R.H.S.


(b) By putting x = 1

2 in L.H.S. we get

L.H.S. = 3 1

2

2

– 8 1

2 – 3

= 3 × 1

4 –

8

2 – 3

=3

4 – 4 – 3

=3

4 – 7

=3 28

4

=25

4

R.H.S.

L.H.S. R.H.S.

Thus equation is not satisfied.

So 1

2 is not the root of the given quadratic equation.


SCHOOL SECTION64

(c) By putting x = 3 in L.H.S. we getL.H.S. = 3 (3)2 – 8 (3) – 3

= 3 (9) – 24 – 3= 27 – 27= 0= R.H.S.


(v) x2 – 4x = 0; x = 0, – 2, 4 (3 marks)Sol. x2 – 4x = 0; x = 0, – 2, 4

(a) By putting x = 0 in L.H.S. we getL.H.S. = (0)2 – 4 (0)

= 0 – 0= 0= R.H.S.


(B) By putting x = – 2 in L.H.S. we getL.H.S. = (– 2)2 – 4 (– 2)

= 4 + 8= 12 R.H.S.

L.H.S. R.H.S.Thus equation is not satisfied.So – 2 is not the root of the given quadratic equation.

(C) By putting x = 4 in L.H.S. we getL.H.S. = (4)2 – 4 (4)

= 16 – 16= 0= R.H.S.


EXERCISE - 2.2 (TEXT BOOK PAGE NO. 38) :4. State whether k is the root of the given equation y2 – (k – 4)y – 4k = 0.

(4 marks)Sol. y2 – (k – 4)y – 4k = 0.

By putting y = k in L.H.S. we getL.H.S. = (k)2 – (k – 4)(k) – 4k

= k2 – (k2 – 4k) – 4k= k2 – k2 + 4k – 4k= 0= R.H.S.

L.H.S. = R.H.S.Thus equation is satisfied.So k is the root of the given quadratic equation.


SCHOOL SECTION 65


11. State whether x =–k2

is the root of the quadratic equation

2x2 + (k – 6) x – 3k = 0 (3 marks)

Sol. 2x2 + (k – 6) x – 3k = 0; x =– k

2

Putting x = – k

2 in L.H.S. we get

L.H.S. =– k – k

2 (k – 6) – 3k2 2

2

=k (k 6k)

2 – 3k4 2

2 2

=k (–k 6k)

– 3k2 2

2 2

=k – k 6k

– 3k2

2 2

=6k

– 3k2

= 3k – 3k= 0= R.H.S.


So – k


EXERCISE - 2.2 (TEXT BOOK PAGE NO. 38) :2. If one root of the quadratic equation x2 – 7x + k = 0 is 4, then find the

value of k. (2 marks)Sol. x2 – 7x + k = 0

x = 4 is the root of given quadratic equation.So it satisfies the given equation.

(4)2 – 7(4) + k = 0 16 – 28 + k = 0 –12 + k = 0

k = 12

3. If one root of the quadratic equation 3y2 – ky + 8 = 0 is 23

, then find the

value of k. (3 marks)Sol. 3y2 – ky + 8 = 0

y = 2

3 is the root of given quadratic equation.

So it satisfies the given equation.

32

3

2

– k2

3 + 8 = 0


SCHOOL SECTION66

3 4

9

–

2k

3 + 8 = 0

4

3 –

2k

3 + 8 = 0

Multiplying throughout by 3, we get,4 – 2k + 24 = 0

– 2k = –28

k =28

2 k = 14


7. Find the value of k if x = 4 is the solution of the equation3x2 + kx – 2 = 0 (3 marks)

Sol. 3x2 + kx – 2 = 0

x = 4 is the solution of given quadratic equation.

Substituting x = 4 in given quadratic equation, it will get satisfied.

3 (4)2 + k (4) – 2 = 0

3 (16)2 + 4k – 2 = 0

48 + 4k – 2 = 0

4k + 46 = 0

4k = – 46

k =46

4

k =23

2

EXERCISE - 2.2 (TEXT BOOK PAGE NO. 38) :5. If one root of the quadratic equation kx2 – 7x + 12 = 0 is 3, then find the

value of k. (2 marks)Sol. kx2 – 7x + 12 = 0

x = 3 is root of given quadratic equation.

So it satisfies the given equation

k(3)2 – 7(3) + 12 = 0 k(9) – 21 + 12 = 0 9k – 9 = 0 9k = 9

k =9

9

k = 1

o Method for solving quadratic equations :(I) Factorisation method

(II) Completing square method

(III) Formula method


SCHOOL SECTION 67

o Method 1 : FACTORIZATIONIn this method, we solve quadratic equations by finding the factors of it.Factors can be found either by splitting the middle term, using the formulaa2 – b2 = (a + b) (a – b) or taking the common terms and writing the factorsin product form.If the given equation ax2 + bx + c = 0 can be written as

(px + q) (mx + n) = 0 (px + q) = 0 or (mx + n) = 0 px = – q or mx = – n

x = – q

p or x = – n

m

– q

p and – n

m are the roots of the given quadratic equation.

EXERCISE - 2.3 (TEXT BOOK PAGE NO. 41) :1. Solve the following quadratic equations by factorization method.(i) x2 – 5x + 6 = 0 (2 marks)Sol. x2 – 5x + 6 = 0

x2 – 3x – 2x + 6 = 0 x(x – 3) – 2(x – 3) = 0 (x – 3) (x – 2) = 0 x – 3 = 0 or x – 2 = 0 x = 3 or x = 2

(ii) x2 + 10x + 24 = 0 (2 marks)Sol. x2 + 10x + 24 = 0

x2 + 6x + 4x + 24 = 0 x(x + 6) + 4(x + 6) = 0 (x + 6) (x + 4) = 0 x + 6 = 0 or x + 4 = 0

x = –6 or x = –4

(iii) x2 – 13x – 30 = 0 (2 marks)Sol. x2 – 13x – 30 = 0

x2 – 15x + 2x – 30 = 0 x (x –15) + 2 (x–15) = 0 (x – 15) (x + 2) = 0 x – 15 = 0 or x + 2 = 0 x = 15 or x = –2

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :9. Solve the following equations by factorization method.(iii) y2 + 8y + 16 = 0 (2 marks)Sol. y2 + 8y + 16 = 0

y2 + 4y + 4y + 16 = 0 y (y + 4) + 4 (y + 4) = 0 (y + 4) (y + 4) = 0 (y + 4)2 = 0

Taking square root on both the sides, we get,y + 4 = 0

y = – 4

y = – 4


SCHOOL SECTION68

(iv) y2 + 3y – 18 = 0 (2 marks)Sol. y2 + 3y – 18 = 0

y2 – 3y + 6y – 18 = 0 y (y – 3) + 6 (y – 3) = 0 (y – 3) (y + 6) = 0 y – 3 = 0 or y + 6 = 0

y = 3 or y = – 6

(v) x2 + 5x + 6 = 0 (2 marks)Sol. x2 + 5x + 6 = 0

x2 + 2x + 3x + 6 = 0 x (x + 2) + 3 (x + 2) = 0 (x + 2) (x + 3) = 0 x + 2 = 0 or x + 3 = 0

x = – 2 or x = – 3

(vi) y2 – 16y + 63 = 0 (2 marks)Sol. y2 – 16y + 63 = 0

y2 – 9y – 7y + 63 = 0 y (y – 9) – 7 (y – 9) = 0 (y – 9) (y – 7) = 0 y – 9 = 0 or y – 7 = 0

y = 9 or y = 7

(ix) y2 – 5y – 24 = 0 (2 marks)Sol. y2 – 5y – 24 = 0

y2 – 8y + 3y – 24 = 0 y (y – 8) + 3 (y – 8) = 0 (y – 8) (y + 3) = 0 y – 8 = 0 or y + 3 = 0

y = 8 or y = – 3

EXERCISE - 2.3 (TEXT BOOK PAGE NO. 41) :1. Solve the following quadratic equations by factorization method.(iv) x2 – 17x + 60 = 0 (2 marks)Sol. x2 – 17x + 60 = 0

x2 – 12x – 5x + 60 = 0 x(x – 12) – 5(x – 12) = 0 (x – 12) (x – 5) = 0 x – 12 = 0 or x – 5 = 0

x = 12 or x = 5

(vii) x2 = 2(11x – 48) (2 marks)Sol. x2 = 2(11x – 48)

x2 = 22x – 96 x2 – 22x + 96 = 0 x2 – 16x – 6x + 96 = 0 x (x – 16) – 6 (x – 16) = 0 (x – 16) (x – 6) = 0 x – 16 = 0 or x – 6 = 0

x = 16 or x = 6


SCHOOL SECTION 69

(viii) 21x = 196 – x2 (2 marks)Sol. 21x = 196 – x2

x2 + 21x – 196 = 0 x2 – 28x – 7x – 196 = 0 x (x + 28) – 7 (x + 28) = 0 (x + 28) (x – 7) = 0 x + 28 = 0 or x – 7 = 0

x = –28 or x = 7

(x) x2 – x – 132 = 0 (2 marks)Sol. x2 – x – 132 = 0

x2 – 12x + 11x – 132 = 0 x (x – 12) + 11 (x – 12) = 0 (x – 12) (x + 11) = 0 x – 12 = 0 or x + 11 = 0

x = 12 or x = –11

(xxiii) x2 – 3 3 x + 6 = 0 (3 marks)Sol. x2 – 3 3 x + 6 = 0

x2 – 2 3 x – 3 x + 6 = 0 x2 – 2 3 x – 3 x + 2 × 3 = 0 x(x – 2 3 ) – 3 (x – 2 3 ) = 0 (x – 2 3 ) (x – 3 ) = 0 x – 2 3 = 0 or x – 3 = 0

x = 2 3 or x = 3

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :18. Solve the following quardratic equations by factorization method.(ii) y2 – 3 = 0 (2 marks)Sol. y2 – 3 = 0

(y)2 – ( 3 )2 = 0

(y + 3 ) (y – 3 ) = 0

y + 3 = 0 or y – 3 = 0

y = – 3 or y = 3

EXERCISE - 2.3 (TEXT BOOK PAGE NO. 41) :

1. Solve the following quadratic equations by factorization method.(xi) 5x2 – 22x – 15 = 0 (3 marks)Sol. 5x2 – 22x – 15 = 0

5x2 – 25x + 3x – 15 = 0 5x(x – 5) + 3(x – 5) = 0 (x – 5) (5x + 3) = 0 x – 5 = 0 or 5x + 3 = 0 x = 5 or 5x = –3

x = 5 or x = –3

5


SCHOOL SECTION70

(xii) 3x2 – x – 10 = 0 (3 marks)Sol. 3x2 – x – 10 = 0

3x2 – 6x + 5x – 10 = 0 3x(x – 2) + 5(x – 2) = 0 (x – 2) (3x + 5) = 0 x – 2 = 0 or 3x + 5 = 0 x = 2 or 3x = –5

x = 2 or x = –5

3

(xiii) 2x2 – 5x – 3 = 0 (3 marks)Sol. 2x2 – 5x – 3 = 0

2x2 – 6x + x – 3 = 0 x(x – 3) + 1(x – 3) = 0 (x – 3) (2x + 1) = 0 x – 3 = 0 or 2x + 1 = 0 x = 3 or 2x = –1

x = 3 or x = –1

2

(xiv) x (2x + 3) = 35 (3 marks)Sol. x (2x + 3) = 35

2x2 + 3x – 35 = 0 2x2 + 10x – 7x – 35 = 0 2x (x + 5) – 7 (x + 5) = 0 (x + 5) (2x – 7) = 0 x + 5 = 0 or 2x – 7 = 0 x = – 5 or 2x = 7

x = – 5 or x = 7

2

(xv) 7x2 + 4x – 20 = 0 (3 marks)Sol. 7x2 + 4x – 20 = 0

7x2 + 14x – 10x – 20 = 0 7x (x + 2) – 10 (x + 2) = 0 (x + 2) (7x – 10) = 0 x + 2 = 0 or 7x – 10 = 0 x = – 2 or 7x = 10

x = – 2 or x = 10

7

(xvi) 10x2 + 3x – 4 = 0 (3 marks)Sol. 10x2 + 3x – 4 = 0

10x2 + 8x – 5x – 4 = 0 2x (5x + 4) – 1 (5x + 4) = 0 (5x + 4) (2x – 1) = 0 5x + 4 = 0 or 2x – 1 = 0 5x = – 4 or 2x = 1

x = –4

5or x =

1

2


SCHOOL SECTION 71

(xvii) 6x2 – 7x – 13 = 0 (3 marks)Sol. 6x2 – 7x – 13 = 0

6x2 – 13x + 6x – 13 = 0 x(6x – 13) + 1(6x – 13) = 0 (6x – 13) (x + 1) = 0 6x – 13 = 0 or x + 1 = 0 6x = 13 or x = –1

x = 13

6or x = –1

(xviii) 3x2 + 34x + 11 = 0 (3 marks)Sol. 3x2 + 34x + 11 = 0

3x2 + 33x + x + 11 = 0 3x (x + 11) + 1 (x + 11) = 0 (x + 11) (3x + 1) = 0 x + 11 = 0 or 3x + 1 = 0 x = – 11 or 3x = – 1

x = – 11 or x = –1

3

(xix) 3x2 – 11x + 6 = 0 (3 marks)Sol. 3x2 – 11x + 6 = 0

3x2 – 9x – 2x + 6 = 0 3x(x – 3) – 2(x – 3) = 0 (x – 3) (3x – 2) = 0 x – 3 = 0 or 3x – 2 = 0 x = 3 or 3x = 2

x = 3 or x = 2

3

(xx) 3x2 – 10x + 8 = 0 (3 marks)Sol. 3x2 – 10x + 8 = 0

3x2 – 6x – 4x + 8 = 0 3x(x – 2) – 4(x – 2) = 0 (x – 2) (3x – 4) = 0 x – 2 = 0 or 3x – 4 = 0 x = 2 or 3x = 4

x = 2 or x = 4

3

(xxi) 2m2 + 19m + 30 = 0 (3 marks)Sol. 2m2 + 19m + 30 = 0

2m2 + 15m + 4m + 30 = 0 m(2m + 15) + 2(2m + 15) = 0 (2m + 15) (m + 2) = 0 2m + 15 = 0 or m + 2 = 0 2m = –15 or m = –2

m = –15

2or m = –2


SCHOOL SECTION72

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :18. Solve the following quardratic equations by factorization method.(i) 4y2 + 4y + 1 = 0 (3 marks)Sol. 4y2 + 4y + 1 = 0

4y2 + 2y + 2y + 1 = 0 2y (2y + 1) + 1 (2y + 1) = 0 (2y + 1) (2y + 1) = 0 (2y + 1)2 = 0

Taking square root on both the sides, we get;2y + 1 = 0

2y = – 1

y =1

2

(iii) 3x2 + 10 = 11x (3 marks)Sol. 3x2 + 10 = 11x

3x2 – 11x + 10 = 0 3x2 – 6x – 5x + 10 = 0 3x (x – 2) – 5 (x – 2) = 0 (x – 2) (3x – 5) = 0 x – 2 = 0 or 3x – 5 = 0 x = 2 or 3x = 5

x = 2 or x = 5

3

(vi) 8x2 – 22x – 21 = 0 (3 marks)Sol. 8x2 – 22x – 21 = 0

8x2 – 28x + 6x – 21 = 0 4x (2x – 7) + 3 (2x – 7) = 0 (2x – 7) (4x + 3) = 0 2x – 7 = 0 or 4x + 3 = 0 2x = 7 or 4x = – 3

x = 7

2or x =

3

4

(vii) 9y2 – 3y – 2 = 0 (3 marks)Sol. 9y2 – 3y – 2 = 0

9y2 – 6y + 3y – 2 = 0 3y (3y – 2) + 1 (3y – 2) = 0 (3y – 2) (3y + 1) = 0 3y – 2 = 0 or 3y + 1 = 0 3y = 2 or 3y = – 1

y = 2

3or y =

1

3

(viii) 5z2 – 3z – 2 = 0 (3 marks)Sol. 5z2 – 3z – 2 = 0

5z2 – 5z + 2z – 2 = 0 5z (z – 1) + 2 (z – 1) = 0 (z – 1) (5z + 2) = 0 z – 1 = 0 or 5z + 2 = 0 z = 1 or 5z = – 2

z = 1 or z =2

5


SCHOOL SECTION 73

(xi) 9x2 + 8 = 22x (3 marks)Sol. 9x2 + 8 = 22x

9x2 – 22x + 8 = 0 9x2 – 18x – 4x + 8 = 0 9x (x – 2) – 4 (x – 2) = 0 (x – 2) (9x – 4) = 0 x – 2 = 0 or 9x – 4 = 0 x = 2 or 9x = 4

x = 2 or x = 4

9

(xii) 6y2 + 17y + 12 = 0 (3 marks)Sol. 6y2 + 1y + 12 = 0

6y2 + 8y + 9y + 12 = 0 2y (3y + 4) + 3 (3y + 4) = 0 (3y + 4) (2y + 3) = 0 3y + 4 = 0 or 2y + 3 = 0 3y = – 4 or 2y = – 3

y = 4

3

or y =

3

2

(iv)19

x2 – 23

– x + 1 = 0 (3 marks)

Sol.1

9x2 –

2

3 – x + 1 = 0

Multiplying throughout by 9, we get;x2 – 6x + 9 = 0

x2 – 3x – 3x + 9 = 0 x (x – 3) – 3 (x – 3) = 0 (x – 3) (x – 3) = 0 (x – 3)2 = 0

Taking square root on both the sides, we get;x – 3 = 0

x = 3

(x) 5t2 – 172

t + 32

= 0 (3 marks)

Sol. 5t2 – 17

2t +

3

2 = 0

Multiplying throughout by 2, we get,10t2 – 17t + 3 = 0

10t2 – 15t – 2t + 3 = 0 5t (2t – 3) – 1 (2t – 3) = 0 (2t – 3) (5t – 1) = 0 2t – 3 = 0 or 5t – 1 = 0 2t = 3 or 5t = 1

t = 3

2or t =

1

5


SCHOOL SECTION74

(ix)1

x 2+ = 2

1x

(3 marks)

Sol.1

x + 2 = 2

1

x x2 = x + 2 x2 – x – 2 = 0 x2 – 2x + x – 2 = 0 x (x – 2) + 1 (x – 2) = 0 (x – 2) (x + 1) = 0 x – 2 = 0 or x + 1 = 0

x = 2 or x = – 1


1. Solve the following quadratic equations by factorization method.

(vi) x + 20x

– 12 = 0 (3 marks)

Sol. x + 20

x – 12 = 0

Multiplying throughout by x, we get,x2 + 20 – 12x = 0

x2 – 12x + 20 = 0 x2 – 10x – 2x + 20 = 0 x(x – 10) – 2(x – 10) = 0 (x – 10) (x – 2) = 0 x – 10 = 0 or x – 2 = 0

x = 10 or x = 2

(ix) 2x – 10x

= 1 (3 marks)

Sol. 2x – 10

x = 1

Multiplying throughout by x, we get, 2x2 – 10 = x 2x2 – x – 10 = 0 2x2 – 5x + 4x – 10 = 0 x(2x – 5) + 2(2x – 5) = 0 (2x – 5) (x + 2) = 0 2x – 5 = 0 or x + 2 = 0 2x = 5 or x = – 2

x = 5

2or x = – 2

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :9. Solve the following equations by factorization method.(ii) 3x – x2 = 0 (3 marks)Sol. 3x – x2 = 0

0 = x2 – 3x x2 – 3x = 0 x (x – 3) = 0 x = 0 or x – 3 = 0

x = 0 or x = 3


SCHOOL SECTION 75

(i) y2 – 3 = 0 (3 marks)Sol. y2 – 3 = 0

y2 – ( 3 )2 = 0 (y + 3 ) (y – 3 ) = 0 y + 3 = 0 or y – 3 = 0

y = – 3 or y = 3

(vii) 9x2 – 16 = 0 (3 marks)Sol. 9x2 – 16 = 0

(3x)2 – (4)2 = 0 3 (x + 4) (3x – 4) = 0 3x + 4 = 0 or 3x – 4 = 0

3x = – 4 or 3x = 4

x = –4

3or x =

4

3

(viii) 49x2 = 36 (3 marks)Sol. 49x2 = 36

49x2 – 36 = 0 (7x)2 – (6)2 = 0 (7x + 6) (7x – 6) = 0 7x + 6 = 0 or 7x – 6 = 0 7x = – 6 or 7x = 6

x = –6

7or x =

6

7


1. Solve the following quadratic equations by factorization method.

(v) m2 – 84x = 0 (3 marks)

Sol. m2 – 84 = 0

(m)2 – 842

= 0

(m)2 – 4 212

= 0

(m2 – 2 212

= 0

m 2 21 m – 2 21 = 0

m 2 21 = 0 or m – 2 21 = 0

m = –2 21 or m = 2 21

(xxii) 7m2 – 84 = 0 (3 marks)Sol. 7m2 – 84 = 0

7(m2 – 12) = 0

m2 –12 =0

7

(m)2 – 122

= 0

(m)2 – 4 32

= 0

(m + 2 3 ) (m – 2 3 ) = 0

m + 2 3 = 0 or m – 2 3 = 0

m = –2 3 or m = 2 3


SCHOOL SECTION76

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :18. Solve the following quardratic equations by factorization method.(v) (2y + 3)2 = 81 (3 marks)Sol. (2y + 3)2 = 81

(2y + 3)2 – 81 = 0 (2y + 3)2 – (9)2 = 0 (2y + 3 + 9) (2y + 3 – 9) = 0 (2y + 12) (2y – 6) = 0 2y + 12 = 0 or 2y – 6 = 0 2y = – 12 or 2y = 6

y = –12

2or y =

6

2

y = – 6 or y = 3

o Method 2 : COMPLETING SQUARE METHOD(Assuming the variable involved in equation to be x)

Step 1 : Check the coefficient of x2, it has to be 1. If not, then make it 1 bydividing each term by the coefficient of x2.

Step 2 : Keep only the constant term on the R.H.S.Step 3 : Find the third term, using the formula :

Third term =

21× coefficient of x

2Step 4 : Add third term on both the sides. Due to this, L.H.S. becomes a perfect square.Step 5 : Express L.H.S. in a square form.Step 6 : Now take square root on both sides and write the square root of

R.H.S. with ' ' sign.Step 7 : Find the value of x and write the solution set.

EXERCISE - 2.4 (TEXT BOOK PAGE NO. 46) :1. Solve the following quadratic equations by completing square.(i) x2 + 8x + 9 = 0 (3 marks)Sol. x2 + 8x + 9 = 0

x2 + 8x = –9 .....(i)

Third term =1

coefficient of x2

2

=1

82

2

= (4)2

= 16Adding 16 to both sides of (i) we get,x2 + 8x + 16 = –9 + 16

(x + 4)2 = 7

(x + 4)2 = 72

Taking square root on both the sides we get,

x + 4 = 7 x = – 4 + 7 x = – 4 + 7 or x = – 4 – 7

– 4 + 7 and – 4 – 7 are the roots of the given quadratic equations.


SCHOOL SECTION 77

(ii) z2 + 6z – 8 = 0 (3 marks)Sol. z2 + 6z – 8 = 0

z2 + 6z = 8 .... (i)

Third term =1

coefficient of z2

2

=1

62

2

= (3)2

= 9Adding 9 to both sides of (i), we get,z2 + 6z + 9 = 8 + 9

(z + 3)2 = 17Taking square root on both the sides we get,

z + 3 = 17

z = –3 + 17

z = –3 + 17 or z = –3 – 17

– 3 + 17 and – 3 – 17 are the roots of the given quadratic equations.

(iii) m2 – 3m – 1 = 0 (3 marks)Sol. m2 – 3m – 1 = 0

m2 – 3m = 1 .... (i)

Third term =1

coefficient of m2

2

=1

– 32

2

=–3

2

2

=9

4

Adding 9

4 to both sides of (i) we get,

m2 – 3m + 9

4=

91

4

3

m –2

2

=4 9

4

3

m –2

2

=13

4


3

m –2

=13

2

m =3 13

2 2


SCHOOL SECTION78

m =3 13

2

m = 3 13

2

or m =

3 – 13

2

3 13

2

and

3 – 13

2 are the roots of the given quadratic equations.

(iv) y2 = 3 + 4y (3 marks)Sol. y2 = 3 + 4y

y2 – 4y = 3 .... (i)

Third term =1

× coefficient of y2

2

=1

× – 42

2

= (– 2)2

= 4Adding 4 to both sides of (i) we get,y2 – 4y + 4 = 3 + 4

(y – 2)2 = 7Taking square root on both the sides we get,

y – 2 = 7

y = 2 + 7

y = 2 + 7 or y = 2 – 7

2 + 7 and 2 – 7 are the roots of the given quadratic equations.

(v) p2 – 12p + 32 = 0 (3 marks)Sol. p2 – 12p + 32 = 0

p2 – 12p = –32 .... (i)

Third term =1

coefficient of p2

2

=1

–122

2

= (– 6)2

= 36Adding 36 to both sides of (i), we get,p2 – 12p + 36 = – 32 + 36

(p – 6)2 = 4Taking square root on both the sides we get,p – 6 = + 2

p = 6 + 2 p = 6 + 2 or p = 6 – 2 p = 8 or p = 4

8 and 4 are the roots of the given quadratic equations.


SCHOOL SECTION 79

(vi) x (x – 1) = 1 (3 marks)Sol. x (x – 1) = 1

x2 – x = 1 .... (i)

Third term =1

coefficient of x2

2

=1

–12

2

=–1

2

2

=1

4

Adding 1

4 to both sides of equation (i) we get,

x2 – x + 1

4= 1 +

1

4

1

x –2

2

=4 1

4

1

x –2

2

=5

4


x =1

25

2

x =1 ± 5

2

x = 1

2 +

5

2or x =

1

2 –

5

2

x = 1 5

2

or x =

1 – 5

2

1 5

2

and

1 – 5


PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :19. Solve the following quadratic equation by completing square.(i) x2 + 3x + 1 = 0Sol. x2 + 3x + 1 = 0

x2 + 3x = – 1 ..... (i)

Third term =2

1coefficient of x

2

=2

13

2


SCHOOL SECTION80

=2

3

2

=9

4

Adding 9

4 to both the sides of (i) we get,

x2 + 3x + 9

4 = – 1 +

9

4

2

3x

2

=4 9

4

2

3x

2

=5

4


x + 3

2=

5

2±

x =3

2+

5

2

x =– 3 ± 5

2

x = – 3 + 5

2or x =

– 3 – 5

2

– 3 + 5

2 and

– 3 – 5

2 are the roots of the given quadratic equation.

(ii) z2 + 4z – 7 = 0 (3 marks)Sol. z2 + 4z – 7 = 0

z2 + 4z = 7 ..... (i)

Third term =1

× coefficient of z2

2

=1

× 42

2

= (2)2

= 4Adding 4 to both the sides of (i) we get,z2 + 4z + 4 = 7 + 4

(z + 2)2 = 11

(z + 2)2 = ( )2

11

Taking square root on both the sides, we get;

z + 2 = ± 11

z = –2 11

z = –2 + 11 or z = –2 – 11

–2 + 11 and –2 – 11 are the roots of the given quadratic equation.


SCHOOL SECTION 81

(iii) n2 + 3n – 4 = 0 (3 marks)Sol. n2 + 3n – 4 = 0

n2 + 3n = 4 ......(i)

Third term =1

× coefficient of n2

2

=1

× 32

2

=3

2

2

=9

4

Adding 9


n2 + 3n + 9

4 = 4 +

9

4

3

n +2

2

=16 + 9

4

3

n +2

2

=25

4Taking square root on both the sides we get,

n + 3

2=

5±

2

n =–3 5

2 2

n = 3

2

–

5

2or n =

3

2

–

5

2

n = 2

2or n = –

8

2 n = 1 or n = – 4

1 and – 4 are the roots of the given quadratic equation.

(iv) m2 = 4 + 5m (3 marks)Sol. m2 = 4 + 5m

m2 – 5m = 4 ..... (i)

Third term =1

× coefficient of m2

2

=1

× –52

2

=–5

2

2

= 25

4


SCHOOL SECTION82

Adding 25


m2 – 5m + 25

4= 4 +

25

4

5

m –2

2

=16 + 25

4

5

m –2

2

=41


5m –

2=

41±

2

m =5 41

2 2

m =5 41

2

m = 5 41

2

+or m =

5 41

2

5+ 41

2 and

5 41

2

are the roots of the given quadratic equation.

(v) p2 – 10p + 5 = 0 (3 marks)

Sol. p2 – 10p + 5 = 0

p2 – 10p = – 5 ..... (i)

Third term =1

× coefficientof p2

2

=1

× –102

2

= (– 5)2

= 25

Adding 25 to both the sides of equation (i), we get;

p2 – 10p + 25 = – 5 + 25

(p – 5)2 = 20

p – 5 = + 20 [Taking square roots on both sides]

p – 5 = + 2 5

p = 5 2 5

p = 5 2 5 or p = 5 – 2 5

5 2 5 and 5 – 2 5 are the roots of the given quadratic equation.

(vi) x (x – 5) = 6 (3 marks)Sol. x (x – 5) = 6

x2 – 5x = 6 ..... (i)


SCHOOL SECTION 83

Third term =1

× coefficientof x2

2

=1

× –52

2

=–5

2

2

=25

4

Adding 25


25x – 5x

42

=25

64

5

x –2

2

=24 + 25

4

5

x –2

2

=49

4Taking square root on both the sides, we get;

x =5 7

2 2

x = 5

2 +

7

2or x =

5

2 –

7

2

x = 12

2or x =

2

2

x = 6 or x = – 1


EXERCISE - 2.4 (TEXT BOOK PAGE NO. 46) :1. Solve the following quadratic equations by completing square.(vii) 3y2 + 7y + 1 = 0 (3 marks)Sol. 3y2 + 7y + 1 = 0

3y2 + 7y = –1Dividing both the sides by 3 we get,

y2 + 7

3y =

–1

3.....(i)

Third term =1


2

=1 7

×2 3

2

=7

6

2

=49

36


SCHOOL SECTION84

Adding 49


y2 + 7

3y +

49

36=

–1 49+

3 36

7

y +6

2

=–12 + 49

36

7

y +6

2

=37

36

Taking square root on both the sides, we get,

y + 7

6=

37

6

y =7 37

–6 6

y =– 7 37

6

y = – 7 37

6

or y =

– 7 – 37

6

– 7 37

6

and

– 7 – 37


(viii) 4p2 + 7 = 12p (3 marks)Sol. 4p2 + 7 = 12p

4p2 – 12p = –7Dividing both the sides by 4 we get,

p2 – 3p = –7

4.... (i)

We make L.H.S. a perfect square by using

Third term =1

× coefficient of p2

2

=1

× – 32

2

=– 3

2

2

=9

4

Adding 9


p2 – 3p + 9

4=

–7

4 +

9

4

3

p –2

2

=2

4


SCHOOL SECTION 85


3p –

2=

2

2

p =3 2

2 2

p =3 2

2

p = 3 2

2

or p =

3 – 2

2

3 2

2

and

3 – 2


(ix) 6m2 + m = 2 (3 marks)Sol. 6m2 + m = 2

Dividing both the sides by 4 we get,

m2 + 1

6m =

1

3.....(i)

Third term =1

× coefficient of m2

2

=1 1

×2 6

2

=1

12

2

=1

144

Adding 1


m2 + 1

6m +

1

144=

1

3 +

1

144

1

m12

2

=48 1

144

1

m12

2

=49


1m

12 =

7

12

m = –1

12 +

7

12or m =

–1

12 –

7

12

m = 6

12or m =

8

12

m = 1

2or m =

2

3

1

2 and

2

3

are the roots of the given quadratic equations.


SCHOOL SECTION86

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :19. Solve the following quadratic equations by completing square :(vii) 2y2 + 5y + 1 = 0 (3 marks)Sol. 2y2 + 5y + 1 = 0

2y2 + 5y = – 1Dividing both the sides by 2 we get,

y2 + 5

2y =

1

2

.....(i)

Third term =1


2

=1 5

×2 2

2

=5

4

2

=25

16

Adding 25


y2 + 5

2y +

25

16=

1

2

+

25

16

5

y +4

2

=–8 + 25

16

5

y +4

2

=17


y + 5

4=

17±

2

y =–5 17

4 4

` y =–5 17

4

y = –5 + 17

4or y =

–5 – 17

4

–5 + 17

4 and

–5 – 17


(viii) 3p2 + 4 = – 7p (3 marks)Sol. 3p2 + 4 = – 7p

3p2 + 7p = – 4Dividing both the sides by 3 we get,

p2 + 7

3p =

4

3

.....(i)

We make L.H.S. a perfect square by using


SCHOOL SECTION 87

Third term =1

× coefficient of p2

2

=1 7

×2 3

2

=7

6

2

=49

36

Adding 49


p2 + 7

3p +

49

36=

4

3

+

49

36

7

p +6

2

=– 48 + 49


p + 7

6=

1

6±

p =–7 1

6 6

p = – 7

6 +

1

6or p = –

7

6 –

1

6

p = – 6

6or p = –

8

6

p = – 1 or p = – 4

3

– 1 and – 4


(ix) 5m2 + m = 3Sol. 5m2 + m = 3

Dividing both the sides by 5 we get,

m2 + 1

5m =

3

5.... (i)

Third term =1

× coefficientof m2

2

=1 1

×2 5

2

=1

10

2

=1

100

Adding 1



SCHOOL SECTION88

m2 + 1

5m +

1

100=

3

5 +

1

100

1

m +10

2

=60 + 1

100

1

m +10

2

=61


m + 1

10=

61±

10

m =–1 61

10 10

m =–1 61

10

m = –1 + 61

10or m =

–1 – 61

10

–1 + 61

10 and

–1 – 61


o Method 3 : FORMULA METHODAn ancient Indian mathematician Shridharacharya arond 1025 A.D. deriveda formula for solving quadratic equations. So this formula is known asShridharacharya fromula for finding the roots of the quadratic equationsax2 + bx + c = 0Derivation of Formula for Formula Method :ax2 + bx + c = 0

Sol. ax2 + bx + c = 0Dividing throught by 'a', we getax bx c

a a a

2

= 0

b

x xa

2 c–

a=

ca

Third Term =1

× coefficient of x2

2

=1 b

×2 a

2

=b

2a

2

=b

4a

2

2

Adding b

4a

2

2 on both the sides we get,

b b

x xa 4a

2

22 =

c b–

a 4a

2

2

b

x2a

2

=– 4ac b

4a

2

2


SCHOOL SECTION 89

b

x2a

2

=b – 4ac

4a

2

2

b

x2a

=b – 4ac

4a

2

2 [Taking square roots on both the sides]

x =b – 4acb

–2a 2a

2

x =– b b – 4ac

2a

2

x = – b b – 4ac

2a

2

is refered as quadratic formula for solving quadratic

equation. The roots of the quadratic equation are denoted as and .

= – b b – 4ac

2a

2

, = – b – b – 4ac

2a

2

Solution set = – b b – 4ac – b – b – 4ac

,2a 2a

2 2

o Steps involved in Formula Method :(Assuming the variable involved in equation to be x)

Step 1 : Write the equation in the standard form.Step 2 : Comparing the given equation with standard form

i.e. ax² + bx + c = 0, write the value of a, b and c.Step 3 : Find the value of b2 - 4ac.

Step 4 : Find the value of x using formula : x =– b b – 4ac

2a

2

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :20. Solve the following quadratic equations using formula.(i) x2 + 3x – 10 = 0 (3 marks)Sol. x2 + 3x – 10 = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = 3, c = – 10b2 – 4ac = (3)2 – 4 (1) (– 10)

= 9 + 40= 49

x =– b ± b -4ac

2a

2

=– 3 ± 49

2 (1)

=–3 ± 7

2

x = –3 + 7

2or x =

–3 – 7

2

x = 4

2or x =

–10

2 x = 2 or x = – 5

2 and – 5 are the roots of given quadratic equation.


SCHOOL SECTION90

EXERCISE - 2.5 (TEXT BOOK PAGE NO. 49) :1. Solve the following quadratic equations by using formula.(i) m2 – 3m – 10 = 0 (3 marks)Sol. m2 – 3m – 10 = 0

Comparing with am2 + bm + c = 0 we have a = 1, b = –3, c = –10b2 – 4ac = (– 3)2 – 4 (1) (–10)

= 9 + 40= 49

m =– b ± b – 4ac

2a

2

=– (– 3) ± 49

2 (1)

=3 7

2

m =3 7

2

or m =

3 – 7

2

m =10

2or m =

–4

2 m = 5 or m = –2

5 and –2 are the roots of the given quadratic equation.

(ii) x2 + 3x – 2 = 0 (3 marks)Sol. x2 + 3x – 2 = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = 3, c = –2b2 – 4ac = (3)2 – 4(1) (–2)

= 9 + 8= 17

x =– b ± b – 4ac

2a

2

=– 3 ± 17

2 (1)

=– 3 ± 17

2

x = – 3 + 17

2or x =

– 3 – 17

2

– 3 + 17

2 and

– 3 – 17


PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :20. Solve the following quadratic equations using formula.(ii) 2x2 + 5x – 2 = 0 (3 marks)Sol. 2x2 + 5x – 2 = 0


= 25 + 16= 41


SCHOOL SECTION 91

x =– b ± b – 4ac

2a

2

=–5 ± 41

2(2)

=–5 ± 41

4

x = –5 + 41

4or x =

–5 – 41

4

–5 + 41

4 and

– 5 – 41


EXERCISE - 2.5 (TEXT BOOK PAGE NO. 49) :1. Solve the following quadratic equations by using formula :(iv) 5m2 – 2m = 2 (3 marks)Sol. 5m2 – 2m = 2

5m2 – 2m – 2 = 0Comparing with ax2 + bx + c = 0 we have a = 5, b = – 2, c = – 2b2 – 4ac = (– 2)2 – 4 (5) (– 2)

= 4 + 40= 44

m =– b b – 4ac

2a

2

=– (– 2) 44

2 (5)

=2 4 11

10

=2 2 11

10

= 2 1 11

10

=1 11

5

x = 1 11

5

or x =

1 – 11

5

1 11

5

and

1 – 11


(v) 7x + 1 = 6x2 (3 marks)Sol. 0 = 6x2 – 7x – 1

Comparing with ax2 + bx + c = 0 we have a = 6, b = – 7, c = – 1b2 – 4ac = (– 7)2 – 4 (6) (– 1)

= 49 + 24= 73


SCHOOL SECTION92

x =– b b – 4ac

2a

2

=– (– 7) 73

2 (6)

=7 73

12

x = 7 73

12

or x =

7 – 73

12

7 73

12

and

7 – 73


(vi) 2x2 – x – 4 = 0 (3 marks)Sol. 2x2 – x – 4 = 0

Comparing with ax2 + bx + c = 0 we have a = 2, b = – 1, c = – 4b2 – ac = (– 1)2 – 4 (2) (– 4)

= 1 + 32= 33

x =– b b – 4ac

2a

2

=– (–1) 33

2 (2)

=1 33

4

x = 1 33

4

or x =

1 – 33

4

1 33

4

and

1 – 33


(vii) 3y2 + 7y + 4 = 0 (3 marks)Sol. 3y2 + 7y + 4 = 0

Comparing with ax2 + bx + c = 0 we have a = 3, b = 7, c = 4b2 – ac = (7)2 – 4 (3) (4)

= 49 – 48= 1

x =– b b – 4ac

2a

2

=– 7 1

2 (3)

=– 7 1

6

y = –7 1

6

or y =

–7 – 1

6

y = – 1 or y = – 4

3

– 1 and – 4



SCHOOL SECTION 93

(viii) 2n2 + 5n + 2 = 0 (3 marks)Sol. 2n2 + 5n + 2 = 0

Comparing with ax2 + bx + c = 0 we have a = 2, b = 5, c = 2b2 – 4ac = (5)2 – 4 (2) (2)

= 25 – 16= 9

n =– b b – 4ac

2a

2

=–5 9

2 (2)

=–5 3

4

n = –5 3

4

or n =

–5 – 3

4

n = – 2

4or n =

– 8

4

n = –1

2or n = – 2

–1


(ix) 7p2 – 5p – 2 = 0 (3 marks)Sol. 7p2 – 5p – 2 = 0

Comparing with ap2 + bp + c = 0 we have a = 7, b = – 5, c = – 2b2 – 4ac = (– 5)2 – 4 (7) (– 2)

= 25 + 56= 81

p =– b b – 4ac

2a

2

=– (–5) 81

2 (7)

=5 9

14

p = 5 9

14

or p =

5 – 9

14

p = 14

14or p =

– 4

14

p = 1 or p = – 2

7

1 and – 2


(x) 9s2 – 4 = – 6s (3 marks)Sol. 9s2 – 4 = – 6s

9s2 + 6s – 4 = 0


SCHOOL SECTION94

Comparing with as2 + bs + c = 0 we have a = 9, b = 6, c = – 4b2 – ac = (6)2 – 4 (9) (– 4)

= 36 + 144= 180

s =– b b – 4ac

2a

2

=– 6 180

2 (9)

=–6 36 5

18

=– 6 6 5

18

= 6 –1 5

18

=–1 5

3

s = –1 5

3

or s =

–1 – 5

3

–1 5

3

and

–1 – 5


(xi) 3q2 = 2q + 8 (3 marks)Sol. 3q2 = 2q + 8

3q2 – 2q – 8 = 0Comparing with aq2 + bq + c = 0 we have a = 3, b = – 2, c = – 8b2 – 4ac = (– 2)2 – 4 (3) (– 8)

= 4 + 96= 100

q =– b b – 4ac

2a

2

=– (– 2) 100

2 (3)

=2 10

6

q = 2 10

6

or q =

2 – 10

6

q = 12

6or q =

– 8

6

q = 2 or q = – 4

3

2 and – 4


(xii) 4x2 + 7x + 2 = 0 (3 marks)Sol. 4x2 + 7x + 2 = 0

Comparing with ax2 + bx + c = 0 we have a = 4, b = 7, c = 2


SCHOOL SECTION 95

b2 – 4ac = (7)2 – 4 (4) (2)= 49 – 32= 17

x =– b b – 4ac

2a

2

=– 7 17

2 (4)

=– 7 17

8

x = – 7 17

8

or x =

– 7 – 17

8

– 7 17

8

and

– 7 – 17


PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :20. Solve the following quadratic equations using formula.(iv) 6m2 – 4m = 3 (3 marks)Sol. 6m2 – 4m = 3

6m2 – 4m – 3 = 0Comparing with am2 + bm + c = 0, we have a = 6, b = – 4, c = – 3b2 – 4ac = (– 4)2 – 4 (6) (– 3)

= 16 + 72= 88

m =– b ± b -4ac

2a

2

=( )– – 4 ± 88

2(6)

=4 ± 4 × 22

12

=4 ± 2 22

12

= 2 2 22

12

=2 ± 22

6

m = 2 + 22

6or m =

2 – 22

6

2 + 22

6 and

2 – 22


(v) 9x + 1 = 4x2 (3 marks)Sol. 9x + 1 = 4x2

0 = 4x2 – 9x – 1 4x2 – 9x – 1 = 0

Comparing with ax2 + bx + c = 0 we have a = 4, b = – 9, c = – 1


SCHOOL SECTION96

b2 – 4ac = (– 9)2 – 4 (4) (– 1)= 81 + 16= 97

x =– b ± b – 4ac

2a

2

=( )– –9 ± 97

2(4)

=9 ± 97

8

x = 9 + 97

8or x =

9 – 97

8

9 + 97

8 and

9 – 97


(vi) 4x2 + x – 5 = 0 (3 marks)Sol. 4x2 + x – 5 = 0


= 1 + 80= 81

x =– b ± b – 4ac

2a

2

=–1 ± 81

2(4)

=–1 ± 9

8

x = –1 + 9

8or x =

–1 – 9

8

x = 8

8or x =

–10

8

x = 1 or x = 5

4

1 and 5

4


(vii) 3y2 + 8y + 5 = 0 (3 marks)Sol. 3y2 + 8y + 5 = 0

Comparing with ay2 + by + c = 0 we have a = 3, b = 8, c = 5b2 – 4ac = (8)2 – 4 (3) (5)

= 64 – 60= 4

y = – b ± b – 4ac

2a

2

= – 8 ± 4

2 (3)

=–8 ± 2

6


SCHOOL SECTION 97

y = –8 + 2

6or y =

–8 – 2

6

y = 6

6

or y =

– 10

6

y = – 1 or y = 5

3

-

– 1 and 5

3


(viii) 2n2 – 11n + 9 = 0 (3 marks)Sol. 2n2 – 11n + 9 = 0

Comparing with an2 + bn + c = 0 we have a = 2, b = – 11, c = 9b2 – 4ac = (– 11)2 – 4 (2) (9)

= 121 – 72= 49

n =– b ± b – 4ac

2a

2

=( )– –11 ± 49

2 (2)

=11±7

4

n = 11+7

4or n =

11 7

4

n = 18

4or n =

4

4

n = 9

2or n = 1

9

2 and 1 are the roots of the given quadratic equation.

(ix) 9p2 – 5p – 4 = 0 (3 marks)Sol. 9p2 – 5p – 4 = 0

Comparing with ap2 + bp + c = 0 we have a = 9, b = – 5, c = – 4b2 – 4ac = (– 5)2 – 4 (9) (– 4)

= 25 + 144= 169

p =– b ± b – 4ac

2a

2

=( )– –5 ± 169

2 (9)

=5 ± 13

18

p = 5 + 13

18or p =

5 – 13

18

p = 18

18or p =

– 8

18

p = 1 or p = 4

9

1 and 4

9



SCHOOL SECTION98

EXERCISE - 2.5 (TEXT BOOK PAGE NO. 49) :1. Solve the following quadratic equations by using formula :

(iii) x2 + x – 1

3 = 0 (3 marks)

Sol. x2 + x – 1

3 = 0

Multiplying throughout by 3 we get,3x2 + x – 1 = 0Comparing with ax2 + bx + c = 0 we have a = 3, b = 1, c = – 1b2 – 4ac = (1)2 – 4 (3) (– 1)

= 1 + 12= 13

x =– b b – 4ac

2a

2

=–1 13

2 (3)

=–1 13

6

x = –1 13

6

or

–1 – 13

6

–1 13

6

and

–1 – 13


PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :20. Solve the following quadratic equations using formula.

(iii) 2x2 +x – 1

5 = 0 (3 marks)

Sol. 2x2 +x – 1

5 = 0

Multiplying throughout by 5 we get,10x2 + x – 1 = 0Comparing with ax2 + bx + c = 0 we have a = 10, b = 1, c = – 1b2 – 4ac = (1)2 – 4 (10) (– 1)

= 1 + 40= 41

x =– b ± b – 4ac

2a

2

=–1 ± 41

2(10)

=–1 ± 41

20

x = –1 + 41

20or x =

–1 – 41

20

–1 + 41

20 and

–1 – 41



SCHOOL SECTION 99

o Nature of roots of a quadratic equation :We know that the roots of the quadratic equation ax2 + bx + c = 0, a 0 are

– b b – 4ac

2a

2

, – b – b – 4ac

2a

2

Here the nature of the roots of the quadratic equation is determined by thevalue of b2 – 4ac, which is called as discriminant of the quadratic equationand it is denoted by (Delta).

(1) If b2 – 4ac > 0 then b – 4ac2 is positive real and therefore roots of equation

are real and unequal.

e.g. – b b – 4ac

2a

2

, – b – b – 4ac

2a

2

(2) If b2 – 4ac = 0 then b – 4ac2 = 0 and therefore roots of equation are real

and equal so = – b

2a and =

– b

2ai.e. both the roots are real and equal, we say that the equation has repeated roots.

(3) If b2 – 4ac < 0 then b – 4ac2 is not real number and therefore equation

does not have real roots.

Note : If b2 – 4ac > 0 and b2 – 4ac is not a perfect square then theroots of the quadratic equation are irrational and occur in pair. Theyare conjugate of each other.

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :5. Find the values of discriminant of each of the following equation.(i) x2 – 3x + 2 = 0 (1 mark)Sol. x2 – 3x + 2 = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = – 3, c = 2 = b2 – 4ac

= (– 3)2 – 4 (1) (2)= 9 – 8= 1

= 1

(ii) 2x2 + x + 1 = 0 (1 mark)Sol. 2x2 + x + 1 = 0

Comparing with ax2 + bx + c = 0 we have a = 2, b = 1, c = 1 = b2 – 4ac

= (– 1)2 – 4 (2) (1)= 1 – 8= – 7

= – 7

(iii) x2 – 6x + 7 = 0 (1 mark)Sol. x2 – 6x + 7 = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = – 6, c = 7 = b2 – 4ac

= (– 6)2 – 4 (1) (7)= 36 – 28= 8

= 8


SCHOOL SECTION100

EXERCISE - 2.6 (TEXT BOOK PAGE NO. 52) :1. Find the value of discriminant of each of the following equations :(i) x2 + 4x + 1 = 0 (1 mark)Sol. x2 + 4x + 1 = 0


= (4)2 – 4 (1) (1)= 16 – 4= 12

= 12

(ii) 3x2 + 2x – 1 = 0 (1 mark)Sol. 3x2 + 2x – 1 = 0

Comparing with ax2 + bx + c = 0 we have a = 3, b = 2, c = – 1 = b2 – 4ac

= (2)2 – 4 (3) (– 1)= 4 + 12= 16

= 16

(iii) x2 + x + 1 = 0 (1 mark)Sol. x2 + x + 1 = 0


= (1)2 – 4 (1) (1)= 1 – 4= – 3

= – 3

(iv) 23 x + 2 2 x – 2 3 = 0 (1 mark)

Sol. 3 x 2 2 x – 2 3 0 2

Comparing with ax2 + bx + c = 0 we have a = 3 , b = 2 2 , c = –2 3 = b2 – 4ac

= 2 2 –4 3 –2 32

= (4 × 2) + (8 × 3)= 8 + 24= 32

= 32

(v) 4x2 + kx + 2 = 0 (1 mark)Sol. 4x2 + kx + 2 = 0

Comparing with ax2 + bx + c = 0 we have a = 4, b = k, c = 2 = b2 – 4ac

= (k)2 – 4 (4) (2)= k2 – 32

= k2 – 32

(vi) x2 + 4x + k = 0 (1 mark)Sol. x2 + 4x + k = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = 4, c = k


SCHOOL SECTION 101

= b2 – 4ac= (4)2 – 4 (1) (k)= 16 – 4k

= 16 – 4k

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :8. Find the value of discrimininant of each of the following equation.

(i) 23 x + 2 2 x – 2 3 = 0 (1 mark)

Sol. 3 x + 2 2 x – 2 32 = 0

Comparing with ax2 + bx + c = 0 we have a = 3 , b = 2 2 , c = 2 3 = b2 – 4ac

= 2 2 4 3 2 3 2

= 4 × 2 + 8 × 3= 8 + 24= 32

= 32

(ii) 4x2 – kx + 2 = 0 (1 mark)Sol. 4x2 – kx + 2 = 0

Comparing with ax2 + bx + c = 0 we have a = 4, b = – k, c = 2 = b2 – 4ac

= (– k)2 – 4 (4) (2)= k2 – 32

= k2 – 32

(iii) x2 + 4x + k = 0 (1 mark)Sol. x2 + 4x + k = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = 4, c = k = b2 – 4ac

= (4)2 – 4 (1)k= 16 – 4k

= 16 – 4k

EXERCISE - 2.6 (TEXT BOOK PAGE NO. 52) :2. Determine the nature of the roots of the following equations from their

discriminants :(i) y2 – 4y – 1 = 0 (1 mark)Sol. y2 – 4y – 1 = 0

Comparing with ay2 + by + c = 0 we have a = 1, b = – 4, c = – 1 = b2 – 4ac

= (– 4)2 – 4 (1) (-1)= 16 + 4= 20

> 0Hence roots of the quadratic equation are real and unequal.

(ii) y2 + 6y – 2 = 0 (1 mark)Sol. y2 + 6y – 2 = 0

Comparing with ay2 + by + c = 0 we have a = 1, b = 6, c = – 2 = b2 – 4ac

= (6)2 – 4 (1) (– 2)


SCHOOL SECTION102

= 36 + 8= 44


(iii) y2 + 8y + 4 = 0 (1 mark)Sol. y2 + 8y + 4 = 0

Comparing with ay2 + by + c = 0 we have a = 1, b = 8, c = 4 = b2 – 4ac

= (8)2 – 4 (1) (4)= 64 -16= 48


(iv) 2y2 + 5y – 3 = 0 (1 mark)Sol. 2y2 + 5y – 3 = 0


= (5)2 – 4 (2) (– 3)= 25 + 24= 49


(v) 3y2 + 9y + 4 = 0 (1 mark)Sol. 3y2 + 9y + 4 = 0


= (9)2 – 4 (3) (4)= 81 – 48= 33


(vi) 2x2 + 5 3 x + 16 = 0 (1 mark)

Sol. 2x2 + 5 3 x + 16 = 0

Comparing with ax2 + bx + c = 0 we have a = 2, b = 5 3 , c = 16 = b2 – 4ac

= (5 3 )2 – 4 (2) (16)= 25 × 3 – 128= 75 – 128= – 53

< 0Hence roots of the quadratic equation are not real.

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :12. Determine the nature of roots of the following equations from their

discriminants.(i) y2 – 5y + 11 = 0 (1 mark)Sol. y2 – 5y + 11 = 0

Comparing with ay2 + by + c = 0 we have a = 1, b = – 5, c = 11


SCHOOL SECTION 103

= b2 – 4ac= (– 5)2 – 4 (1) (11)= 25 – 44= – 19

< 0Hence roots of the quadratic equation are not real.

(ii) y2 + 6y + 9 = 0 (1 mark)Sol. y2 + 6y + 9 = 0


= (6)2 – 4 (1) (9)= 36 – 36= 0

= 0Hence roots of the quadratic equation are real and equal.

(iii) y2 + 8y + 5 = 0 (1 mark)Sol. y2 + 8y + 5 = 0


= (8)2 – 4 (1) (5)= 64 – 20= 44


(iv) 2y2 – 7y – 3 = 0 (1 mark)Sol. 2y2 – 7y – 3 = 0

Comparing with ay2 + by + c = 0 we have a = 2, b = – 7, c = – 3 = b2 – 4ac

= (– 7)2 – 4 (2) (– 3)= 49 + 24= 73


(v) 2y2 + 11y – 7 = 0 (1 mark)Sol. 2y2 + 11y – 7 = 0


= (11)2 – 4 (2) (– 7)= 121 + 56= 177


(vi) x2 + 3 2 x – 8 = 0 (1 mark)

Sol. x2 + 3 2 x – 8 = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = 3 2 , c = – 8 = b2 – 4ac

= (3 2 )2 – 4 (1) (– 8)= 9 × 2 + 32


SCHOOL SECTION104

= 18 + 32= 50


EXERCISE - 2.6 (TEXT BOOK PAGE NO. 52) :3. Find the value of k for which given equation has real and equal roots :(i) (k – 12)x2 + 2 (k – 12)x + 2 = 0 (3 marks)Sol. (k – 12)x2 + 2 (k – 12)x + 2 = 0

Comparing with ax2 + bx + c = 0 we have a = k – 12, b = 2 (k – 12), c = 2 = b2 – 4ac

= [2 (k – 12)]2 – 4 (k -12) (2)= (2k – 24)2 – 8 (k – 12)= 4k2 – 96k + 576 – 8k + 96= 4k2 – 104k + 672

The roots of given equation are real and equal. must be zero. 4k2 – 104k + 672 = 0 4 (k2 – 26k + 168) = 0 k2 – 14k – 12k + 168 = 0 k (k – 14) – 12 (k – 14)= 0 (k – 14) (k – 12) = 0 k – 14 = 0 or k – 12 = 0

k = 14 or k = 12

(ii) k2x2 – 2 (k – 1)x + 4 = 0 (3 marks)Sol. k2x2 – 2 (k – 1)x + 4 = 0

Comparing with ax2 + bx + c = 0 we have a = k2, b = – 2 (k – 1), c = 4 = b2 – 4ac

= [– 2 (k – 1)]2 – 4 (k2) (4)= (– 2k + 2)2 – 16k2

= 4k2 – 8k + 4 – 16k2

= – 12k2 – 8k + 4 The roots of given equation are real and equal. must be zero. – 12k2 – 8k + 4 = 0 – 4 (3k2 + 2k – 1) = 0

3k2 + 3k – k – 1 =-

0

4 3k (k + 1) – 1 (k + 1) = 0 (k + 1) (3k – 1) = 0 k + 1 = 0 or 3k – 1 = 0 k = – 1 or 3k = 1

k = – 1 or k = 1

3

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :13. Find m, if the roots of the quadratic equation (m – 1) x2 – 2 (m – 1) x + 1 =0

has real and equal roots. (3 marks)Sol. (m – 1) x2 – 2 (m – 1) x + 1 = 0

Comparing with ax2 + bx + c = 0 we have a = m – 1, b = – 2 (m – 1), c = 1 = b2 – 4ac

= [– 2 (m – 1)]2 – 4 (m – 1) (1)


SCHOOL SECTION 105

= (– 2m + 2)2 – 4m + 4= 4m2 – 8m – 4m + 4= 4m2 – 12m + 8

The roots of the given equation are real and equal. must be zero. 4m2 – 12m + 8 = 0 4 (m2 – 3m + 2) = 0

m2 – 3m + 2 =0

4 m2 – 2m – m + 2 = 0 m (m – 2) – 1 (m – 2) = 0 (m – 2) (m – 1) = 0 m – 2 = 0 or m – 1 = 0

m = 2 or m = 1

14. Find c, if the roots of the quadratic equation x2 – 2 (c + 1) x + c2 = 0 hasreal and equal roots. (3 marks)

Sol. x2 – 2 (c + 1) x + c2 = 0Comparing with Ax2 + Bx + C = 0 we have A = 1, B = – 2 (c + 1), C = c2

= B2 – 4AC= [– 2 (c + 1)]2 – 4 (1) (c2)= (– 2)2 (c + 1)2 – 4c2

= 4 (c2 + 2c + 1) – 4c2

= 4c2 + 8c + 4 – 4c2

= 8c + 4 The roots of the given equation are real and equal. must be zero. 8c + 4 = 0 8c = – 4

c =– 4

8

c =–1

2

o Relation between the roots and coefficients :If a and b are the roots of the quadratic equation ax2 + bx + c = 0

then + = – b

aand ab =

c

aLet us verify the above relations.We have the roots of the quadratic equation ax2 + bx + c = 0

– b b – 4ac

2a

2

and – b – b – 4ac

2a

2

+ =– b b – 4ac – b – b – 4ac

2a 2a

2 2

=–2b

2a

=– b

a

Sum of the roots = – b

a =

– coefficient of x

coefficient of x2


SCHOOL SECTION106

Also,

=– b b – 4ac – b – b – 4ac

2a 2a

2 2

= – b – b – 4ac

4a

22 2

2

=b – (b – 4ac)

4a

2 2

2

=b – b 4ac

4a

2 2

2

=4ac

4a2

=c

a

Product of the roots = c

a =

constant term

coefficient of x2

EXERCISE - 2.7 (TEXT BOOK PAGE NO. 54) :1. If one root of the quadratic equation kx2 – 5x + 2 = 0 is 4 times the

other, find k. (3 marks)Sol. kx2 – 5x + 2 = 0

Comparing with ax2 + bx + c = 0 we have a = k, b = – 5, c = 2Let and be the roots of given quadratic equation. = 4 [Given]

+ = - b

a=

( 5)

k

=

5

k

4 + =5

k

5 =5

k

=5

k ×

1

5

=1

k

Also = a

c=

2

k

4. =2

k[ = 4]

42 =2

k

4 1

k

2

= 2

k

1

k

4 × 1

k2 =2

k

4 × 1

2=

k

k

2

k = 2


SCHOOL SECTION 107

2. Find k, if the roots of the quadratic equation x2 + kx + 40 = 0 are in theratio 2 : 5. (3 marks)

Sol. x2 + kx + 40 = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = k, c = 40Let a and b be the roots of given quadratic equation.Ratio of to is 2 : 5 [Given]Considering common multiple as m. = 2m and = 5m

+ = – b

a=

– k

1 = – k

2m + 5m = – k [ = 2m and = 5m] 7m = – k k = – 7m ......(i)

Also . = c

=

40

1 = 40

2m × 5m = 40 [ = 2m and = 5m] 10m2 = 40

m2 =40

10 m2 = 4

Taking square root on both the sides we get,m = + 2

k = – 7m [From (i)] k = – 7 (2) or k = – 7 (– 2) [ m = + 2]

k = – 14 or k = 14

3. Find k, if one of the roots of the quadratic equation kx2 – 7x + 12 = 0 is 3.(3 marks)

Sol. kx2 – 7x + 12 = 0Comparing with ax2 + bx + c = 0 we have a = k, b = – 7, c = 12Let and be the roots of given quadratic equation. = 3 [Given]

+ = – b

a=

– – 7

k =

7

k

3 + =7

k[ a = 3]

=7

k – 3 ......(i)

Also . =c

a =

12

k

3 × =12

k[ a = 3]

=12

k ×

1

3

=4

k..... (ii)

7

k – 3 =

4

k[From (i) and (ii)]

Multiplying throughout by k we get,

Alternative method : 3 is one of the root of the

quadratic equation,kx2 – 7x + 12 = 0

x = 3 satisfies the equation,Substituting x = 3 in equationwe get,k (3)2 – 7 (3) + 12 = 0

9k – 21 + 12 = 0 9k – 9 = 0 9k = 9

k =9

9

k = 1


SCHOOL SECTION108

7 – 3k = 4 – 3k = 4 – 7 – 3k = – 3

k =– 3

– 3

k = 1

4. If the roots of the equation x2 + px + q = 0 differ by 1, prove that p2 = 1 + 4q.(3 marks)

Proof : x2 + px + q = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = p, c = qLet and be the roots of given quadratic equation. – = 1 ......(i) [Given]

+ = b

a

=

p

1

= – p ......(ii)

Also = c

a =

q

1......(iii)

We know that,( – )2 = ( + )2 – 4

(1)2 = (– p)2 – 4 (q) 1 = p2 – 4q 1 + 4q = p2

p2 = 1 + 4qHence proved.

5. Find k, if the sum of the roots of the quadratic equation 4x2 + 8kx + k + 9 = 0is equal to their product. (3 marks)

Sol. 4x2 + 8kx + k + 9 = 0Comparing with ax2 + bx + c = 0 we have a = 4, b = 8k, c = k + 9Let and be the roots of given quadratic equation. + = . ......(i) [Given]

+ = – b

a =

– 8k

4 = – 2k ......(ii)

Also =c

a =

k + 9

4......(iii)

+ = . [From (i)]

– 2k =k + 9

4[From (ii) and (iii)]

– 2k × 4 = k + 9 – 8k = k + 9 – 8k – k = 9 – 9k = 9

k =9

9 k = – 1

k = – 1

6. If and are the roots of the equation x2 – 5x + 6 = 0, find

(i) 2 + 2 (ii) + (4 marks)

Sol. x2 – 5x + 6 = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = -5, c = 6 and be the roots of given quadratic equation. [Given]


SCHOOL SECTION 109

+ = –

=( )– –5

1 = 5 ......(i)

Also . =c

a =

6

1 = 6 ......(ii)

(i) 2 + 2 = ?We know that,

2 + 2 = ( + )2 – 2,= (5)2 – 2 (6) .....[From (i) and (ii)]= 25 – 12= 13 ..... (iii)

2 + 2 = 13

(ii) + = ?

+ =

+2 2

+ =

13

6

23. If and are the roots of the equation 4x2 – 5x + 2 = 0 find the equationwhose roots are.

(i) + 3 and 3 + (ii) and

(iii)2 and

2

(iv) + 1

and + 1 (5 marks)

Sol. 4x2 – 5x + 2 = 0Comparing with ax2 + bx + c = 0 we have a = 4, b = – 5, c = 2 and are the roots of given quadratic equation

Sum of roots = + = – b

a =

– (–5)

4 =

5

4

Product of roots = . = c

a =

2

4 =

1

2

Now we have = + = 5

4 and . =

1

2(i) + 3 and 3 + are the roots of quadratic equation

Sum of roots = + 3 + 3 + = 4 + 4= 4 ( + )

= 4 × 5

4

5

4

= 5 2 + 2 = ( + )2 – 2.

=5

4

2

× 2 × 1

2

=25

– 116

=25 – 16

16

2 + 2 =9

16


SCHOOL SECTION110

Product of roots= ( + 3) × (3 + )= 32 + + 9 + 32

= 32 + 32 + 10

= 3 (2 + 2) + 10 × 1

2

1

2

= 3 × 9

16 + 5

9

16

2 2

=27

516

=27 80

16

=107

16We know that,x2 – (Sum of roots) x + Product of roots = 0

x2 – 5x + 107

16 = 0

Multiplying throughout by 16 we get,16x2 – 80x + 107 = 0

The required equation is 16x2 – 80x + 107 = 0

(ii) and

are the roots of quadratic equation

Sum of roots =

=2 2

=

91612

9 1and .

16 2

2 2

=9 2

16 1

=9

8

Product of roots=

= 1We know that,x2 – (Sum of roots)x + Product of roots = 0

x2 – 9

8x + 1 = 0


The required quadratic equation is 8x2 – 9x + 8 = 0


SCHOOL SECTION 111

(iii)2 and

2


3 + 3 = ( + )3 – 3 . ( + )

=125 15

–64 8

=125 – 120

64

3 + 3 =5

64

Sum of roots = 2 2

=3 3

=

56412

5 1and .

64 2

2 2

=5 2

64 1

=5

32

Product of roots = 2 2

= .

=1

2

1.

2

We know that,x2 – (Sum of roots) x + Product of roots = 0

x2 – 5

32x +

1

2 = 0

Multiplying throughout by 32 we get,

The required quadratic equation is 32x2 – 5x + 16 = 0

(iv) + 1

and +

1


Sum of roots =1 1

=1 1

=

=

=

55 4

142

5 1and .

4 2

=5 5 2

4 4 1


SCHOOL SECTION112

=5 10

4 4

=15

4

Product of roots =1 1

=1

..

=1

.. .

2 2

=

91 116

1 122 2

1 9and

2 16

2 2

=1 9 2 2

12 16 1 1

=1 9 2

2 8 1

=4 9 16

8

=29

8We know that,x2 – (Sum of roots) x + Product of roots = 0

x2 – 15

4x +

29

8 = 0


The required quadratic equation 8x2 – 30x + 29 = 0.

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 165) :17. If and are the roots of equation ax2 + bx + c = 0. find the value of

+α ββ α . (3 marks)

Sol. ax2 + bx + c = 0 and are the roots of given quadratic equation.We know that,

+ = – b

a and . =

c

a

Now, β

+ =

+

.

2 2

=( )+ – 2 .

.

2

=

– b c– 2 ×

a aca

2


SCHOOL SECTION 113

=b 2c c

– ÷a a a

2

2

=b – 2ac a

×a c

2

2

=b – 2ac

ac

2

+βα

β α =2b 2ac

ac


7. If one root of the quadratic equation kx2 – 20x + 34 = 0 is 5 – 2 2 , find k.

(3 marks)Sol. kx2 – 20x + 34 = 0

Comparing with ax2 + bx + c = 0 we have a = k, b = – 20, c = 34Let and be the roots of given quadratic equation.

= 5 – 2 2 [Given]

= 5 + 2 2

+ = – b

a =

( )– – 20

k =

20

k

5 – 2 2 + 5 + 2 2 =20

k[ = 5 – 2 2 and = 5 + 2 2 ]

10 =20

k

k =20

10

k = 2


15. If one root of the equation x2 – 10x + 2k = 0 is 5 – 3 , find k. (3 marks)

Sol. x2 – 10x + 2k = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = – 10, c = 2kLet and be the roots of given quadratic equation.

= 5 – 3 [Given]

= 5 + 3

. =c

a =

2k

1 = 2k

(5 – 3 ) (5 + 3 ) = 2k [ = 5 – 3 and = 5 + 3 ]

(5)2 – ( 3 )2 = 2k

25 – 3 = 2k 22 = 2k

22

2= k

k = 11


SCHOOL SECTION114

16. If one root of the quadratic equation x2 + 6x + k = 0 is h +2 6 , find h and k.(3 marks)

Sol. x2 + 6x + k = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = 6, c = kLet and be the roots of given quadratic equation.

= h + 2 6 [Given]

= h – 2 6

+ =– b

a =

– 6

1 = – 6

h + 2 6 + h – 2 6 = – 6 [ = h + 2 6 and = h – 2 6 ] 2h = – 6

h =– 6

2 h = – 3

. =c

a =

k

1 = k

(h + 2 6 ) (h – 2 6 )= k [ = h + 2 6 and = h – 2 6 ]

(h)2 – (2 6 )2 = k (– 3)2 – 4 × 6 = k [ h = – 3] 9 – 24 = k k = – 15

h = – 3 and k = – 15

o To form quadratic equation if its roots are given :If and are the roots of the quadratic equation in variable x then,x = and x =

x – = 0 and x – = 0 (x – ) (x – ) = 0 x2 – ( + ) x + = 0

i.e. x2 – (Sum of the roots) x + Product of the roots = 0

EXERCISE - 2.8 (TEXT BOOK PAGE NO. 56) :1. Form the quadratic equation if its roots are(i) 5 and – 7 (2 marks)Sol. The roots of the quadratic equation are 5 and – 7.

Let = 5 and = 7 + = 5 + (– 7) = 5 – 7 = – 2

and . = 5 × – 7 = – 35We know that,x2 – ( + )x + . = 0

x2 – (– 2)x + (– 35) = 0 x2 + 2x – 35 = 0

The required quadratic equation is x2 + 2x – 35 = 0

(ii)12

and – 34

(2 marks)

Sol. The roots of the quadratic equation are 1

2 and

– 3

4

Let = 1

2 and =

– 3

4


SCHOOL SECTION 115

+ = 1

2 +

3

4

=2 – 3

4 =

–1

4

and . = 1

2 ×

– 3

4=

– 3

8We know that,x2 – ( + )x + a.b = 0

x2 – 1

4

x + 3

8

= 0

x2 + x

4 –

3

8= 0

Multiplying throughout by 8 we get,8x2 + 2x – 3 = 0

The required quadratic equation is 8x2 + 2x – 3 = 0

(iii) – 3 and – 11 (2 marks)Sol. The roots of the quadratic equation are -3 and -11.

Let = – 3 and = – 11 + = – 3 + (– 11) = – 3 – 11 = – 14

and . = – 3 × – 11 = 33We know that,x2 – ( + )x + .= 0

x2 – (– 14)x + 33 = 0 x2 + 14x + 33 = 0

The required quadratic equation is x2 + 14x + 33 = 0

(iv) – 2 and 112

(2 marks)

Sol. The roots of the quadratic equation are – 2 and 11

2.

Let = – 2 and =11

2

+ = – 2 + 11

2 =

– 4 + 11

2 =

7

2

and . = – 2 × 11

2 = – 11

We know that,x2 – ( + )x + . = 0

x2 – 7

2x + (– 11) = 0

x2 – 7

2x – 11 = 0

Multiplying throughout by 2 we get,2x2 – 7x – 22 = 0

The required quadratic equation is 2x2 – 7x – 22 = 0

(v)12

and –12

(3 marks)


2 and

–1

2.


SCHOOL SECTION116

Let = 1

2 and =

–1

2

+ = 1

2 +

1

2

=1

2

1–

2 = 0

and . = 1

2 ×

1

2

=

1

4

-

We know that,x2 – ( + )x + . = 0

x2 – 0x + 1

4

= 0

x2 – 1

4= 0

Multiplying throughout by 4 we get,4x2 – 1 = 0

The required quadratic equation is 4x2 – 1 = 0

(vi) 0 and – 4 (2 marks)Sol. The roots of the quadratic equation are 0 and – 4.

Let = 0 and = – 4 + = 0 + (– 4) = – 4

and . = 0 × – 4 = 0We know that,x2 – ( + )x + . = 0

x2 – (– 4)x + 0 = 0 x2 + 4x = 0

The required quadratic equation is x2 + 4x = 0


21. From the quadratic equation whose roots are.(i) 3 and 10 (2 marks)Sol. The roots of the quadratic equation are 3 and 10

Let = 3 and = 10 + = 3 + 10 = 13

. = 3 × 10 = 30We know that,x2 – ( + ) x + . = 0

x2 – 13x + 30 = 0

The required quadratic equation is x2 – 13x + 30 = 0

(ii)34

and –23

(3 marks)


4 and

– 2

3

Let = 3

4 and =

– 2

3

+ = 3 –2

4 3

= 3 2

–4 3

= 9 – 8

12 =

1

2


SCHOOL SECTION 117

. = 3 –2

4 3 =

–1

2We know that,

x2 – ( + ) x + . = 0

x2 – 1

12x +

1–

2

= 0

x2 – 1

12x –

1

2= 0

Multiplying throughout by 12 we get,12x2 – x – 6 = 0

The required quadratic equation is 12x2 – x – 6 = 0

(iii) – 5 and 9 (2 marks)Sol. The roots of the quadratic equation are – 5 and 9

Let = – 5 and = 9 + = – 5 + 9 = 4

. = – 5 × 9 = – 45We know that,

x2 – ( + ) x + . = 0 x2 – 4x + (– 45) = 0 x2 – 4x – 45 = 0

The required quadratic equation is x2 – 4x – 45 = 0

(iv) – 3 and 52

(3 marks)

Sol. The roots of the quadratic equation are – 3 and 5

2

Let = – 3 and = 5

2

+ = – 3 + 5

2 =

–6 5

2

=

–1

2

. = – 3 × 5

2 =

–15

2We know that,

x2 – ( + ) x + . = 0

x2 – –1

2 x –

15

2= 0

Multiplying throughout by 2 we get,2x2 + x – 15 = 0

The required quadratic equation is 2x2 + x – 15 = 0

(v)34

and –34

(3 marks)


4 and

–3

4

Let = 3

4 and =

– 3

4


SCHOOL SECTION118

+ = 3 –3

4 4

= 3 3

–4 4

= 0

. = 3 –3

4 4 =

– 9

16We know that,

x2 – ( + ) x + . = 0

x2 – (0) x + –9

16

= 0

x2 – 0 – 9

16= 0

x2 – 9

16= 0

Multiplying throughout by 16 we get,16x2 – 9 = 0

The required quadratic equation is 16x2 – 9 = 0

(vi) 0 and – 6 (2 marks)Sol. The roots of the quadratic equation are 0 and – 6

Let = 0 and = – 6 + = 0 + (– 6) = 0 – 6 = – 6

. = 0 × – 6 = 0We know that,

x2 – ( + ) x + . = 0 x2 – (– 6) x + 0 = 0 x2 + 6x = 0

The required quadratic equation is x2 + 6x = 0.

EXERCISE - 2.8 (TEXT BOOK PAGE NO. 56) :2. Form the quadratic equation if its one of the root is

(i) 3 – 2 5 (3 marks)

Sol. If one of the root of the quadratic equation is 3 – 2 5 ,

then the other root is 3 + 2 5

= 3 – 2 5 and = 3 + 2 5

+ = 3 – 2 5 + 3 + 2 5 = 6

and . = 3 – 2 5 × 3 + 2 5

= (3)2 – 2 52

= 9 – 4 × 5= 9 – 20= – 11

We know that,x2 – ( + )x + . = 0

x2 – 6x + (– 11) = 0 x2 – 6x – 11 = 0


(ii) 4 – 3 2 (3 marks)Sol. If one of the root of the quadratic equation is 4 – 3 2 ,

then the other root is 4 – 3 2


SCHOOL SECTION 119

= 4 – 3 2 and = 4 + 3 2

+ = 4 – 3 2 + 4 + 3 2 = 8

and . = 4 – 3 2 × 4 + 3 2

= (4)2 – 3 22

= 16 – 9 × 2= 16 – 18= – 2

We know that,x2 – ( + )x + . = 0

x2 – 8x + (– 2) = 0 x2 – 8x – 2 = 0


(iii) +2 3 (3 marks)

Sol. If one of the root of the quadratic equation is 2 + 3 ,then the other root is 2 – 3

Let = 2 + 3 and = 2 – 3

+ = 2 3 2 – 3 = 2 2

and . = 2 3 × 2 – 3

= 2 – 32 2

= 2 – 3= – 1

We know that,x2 – ( + )x + . = 0

x2 – 2 2 x –1 = 0

x2 – 2 2 x – 1 = 0

The required quadratic equation is x2 – 2 2 x – 1 = 0

(iv) 2 3 – 4 (3 marks)

Sol. If one of the root of the quadratic equation is 2 3 – 4 ,

then the other root is 2 3 + 4

Let = 2 3 – 4 and = 2 3 + 4

+ = 2 3 – 4 + 2 3 + 4 = 4 3

and . = 2 3 – 4 + 2 3 + 4

= 2 32 – (4)2

= (4 × 3) – 16= 12 – 16= – 4

We know that,x2 – ( + )x + . = 0

x2 – 4 3 x + (– 4) = 0

x2 – 4 3 x – 4 = 0

The required quadratic equation is x2 – 4 3 x – 4 = 0.


SCHOOL SECTION120

(v) 2 + 5 (3 marks)

Sol. If one of the root of the quadratic equation is 2 + 5 , then the otherroot is 2 – 5 .

Let = 2 + 5 and = 2 – 5

+ = 2 + 5 + 2 – 5 = 4

and . = 2 + 5 × 2 – 5

= (2)2 – 52

= 4 – 5= – 1

We know that,x2 – ( + )x + . = 0

x2 – 4x + (-1) = 0 x2 – 4x – 1 = 0 The required quadratic equation is x2 – 4x – 1 = 0

(vi) 5 – 3 (3 marks)

Sol. If one of the root of the quadratic equation is 5 – 3 , then the other

root is 5 + 3 .

Let = 5 – 3 and = 5 + 3

+ = 5 – 3 + 5 + 3 = 2 5

and . = 5 – 3 × 5 + 3

= 52

– ( 3 )2

= 5 – 3= 2

We know that,x2 – ( + )x + . = 0

x2 – 2 5 x + 2 = 0

The required quadratic equation is x2 – 2 5 x – 2 = 0

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :22. Form the quadratic equation whose one of the root is.

(i) 1 – 3 5 (3 marks)

Sol. The one of the root of the quadratic equation is 1 – 3 5

then the other root is 1 + 3 5

= 1 – 3 5 and = 1 + 3 5

+ = 1 – 3 5 + 1 + 3 5 = 2

. = 1 – 3 5 1 3 5

= (1)2 – 3 52

= 1 – 9 × 5= 1 – 45= – 44


SCHOOL SECTION 121

We know that,x2 – ( + ) x + . = 0

x2 – 2x + (– 44) = 0 x2 – 2x – 44 = 0

The required quadratic equation is x2 – 2x – 44 = 0.

(ii) 3 – 2 3 (3 marks)

Sol. If one of the root of the quadratic equation is 3 – 2 3 then the other root

is 3 + 2 3

Let = 3 – 2 3 and = 3 + 2 3 + =3 – 2 3 + 3 + 2 3 = 6

. = 3 – 2 3 × 3 + 2 3

= (3)2 – 2 32

= 9 – 12= – 3

We know that,x2 – ( + ) x + . = 0

x2 – 6x + (– 3) = 0 x2 – 6x – 3 = 0


(iii) 3 – 7 (3 marks)

Sol. If one of the root of the quadratic equation is 3 – 7 then the other root

is 3 + 7

Let = 3 – 7 and = 3 + 7

+ = 3 – 7 + 3 + 7

= 2 3

. = 3 – 7 3 + 7

= 3 – 72 2

= 3 – 7= – 4

We know that,x2 – ( + ) x + . = 0

x2 – 2 3 + (– 4) = 0

x2 – 2 3 – 4 = 0


(iv) 4 2 – 3 (3 marks)

Sol. If one of the root of the quadratic equation is 4 2 – 3 then the other root

is 4 2 + 3

Let = 4 2 – 3 and = 4 2 + 3


SCHOOL SECTION122

+ = 4 2 – 3 + 4 2 + 3= 8 2

. = 4 2 – 3 × 4 2 + 3

= 4 22 – (3)2

= (16 × 2) – 9= 32 – 9= 23

We know that,x2 – ( + ) x + . = 0

x2 – 8 2 x + 23 = 0

The required quadratic equation is x2 – 8 2 x + 23 = 0

(v) 2 + 3 5 (3 marks)

Sol. If one of the root of the quadratic equation is 2 + 3 5 then the other root

is 2 – 3 5Let = 2 + 3 5 and = 2 – 3 5

+ = 2 + 3 5 + 2 – 3 5

= 4

. = 2 + 3 5 2 – 3 5

= (2)2 – 3 52

= 4 – (9 × 5)= 4 – 45= – 41

We know that,x2 – ( + ) x + . = 0

x2 – 4x + (– 41) = 0 x2 – 4x – 41 = 0


(vi) 7 – 2 (3 marks)

Sol. The one of the root of the quadratic equation is 7 – 2 then the other

root is 7 + 2

= 7 – 2 and = 7 + 2

+ = 7 – 2 + 7 + 2 = 2 7

. = 7 – 2 7 + 2

= 7 – 22 2

= 7– 2= 5

We know that,x2 – ( + ) x + . = 0

x2 – 2 7 + 5 = 0



SCHOOL SECTION 123


3. If the sum of the roots of the quadratic is 3 and sum of their cubes is63, find the quadratic equation. (4 marks)

Sol. Let and be the roots of a quadratic equation. + = 3 [Given]

and3 + 3 = 63We know that,x2 – ( + )x + . = 0 .......(i)

Also, 3 + 3 = ( + )3 – 3 . ( + ) 63 = (3)3 – 3 . (3) [ + = 3 and 3 + 3 = 63] 63 = 27 – 9 . 9 . = 27 – 63 9 . = – 36

. =– 36

9 . = – 4 x2 – ( + )x + . = 0 [From (i)] x2 – 3x + (– 4) = 0 [ + = 3 and . = – 4] x2 – 3x – 4 = 0



25. If + = 5 and 3 + 3 = 35, find a quadratic equation whose roots are and . (4 marks)

Sol. and are the roots of a quadratic equation + = 5 [Given]

3 + 3 = 35We know that,

x2 – ( + ) x + . = 0 ......(i)Also, 3 + 3 = ( + )3 – 3 . ( + )

35 = (5)3 – 3 . (5) [ + = 5 and 3 + 3 = 35] 35 = 125 – 15 . 15 . = 125 – 35 15 . = 90

. =90

15 . = 6 x2 – ( + ) x + . = 0 [From (i)] x2 – 5x + 6 = 0

The required quadratic equation is x2 – 5x + 6 = 0.

4. If the difference of the roots of the quadratic equation is 5 and thedifference of their cubes is 215, find the quadratic equation. (4 marks)

Sol. Let and be the roots of a quadratic equation. – = 5 and3 – 3 = 215 [Given]

We know that,x2 – ( + )x + . = 0 .......(i)


SCHOOL SECTION124

Also, 3 – 3 = ( – )3 + 3 . ( – ) 215 = (5)3 + 3 . (5) [ + = 5 and 3 – 3 = 215] 215 = 125 + 15 . 215 – 125 = 15 . 90 = 15 .

. =90

15 . = 6

Now, – 2 = ( + )2 – 4. (5)2 = ( + )2 – 4 (6) [ . = 5 and . = 6] 25 = ( + )2 – 24 25 + 24 = ( + )2

( + )2 = 49Taking square root on both the sides, we get;

+ = + 7 x2 – (+ )x + . = 0 [From (i)] x2 – (7)x + 6 = 0 or x2 – (– 7)x + 6 = 0 x2 – 7x + 6 = 0 or x2 + 7x + 6 = 0

The required quadratic equation is x2 – 7x + 6 = 0 or x2 + 7x + 6 = 0.


24. If the difference of the roots of the quadratic equation is 3 and differencebetween their cubes is 189, find the quadratic equation. (4 marks)

Sol. Let and be the roots of a quadratic equation – = 3 and 3 – 3 = 189 [Given]

We know that,x2 – ( + ) x + . = 0 ......(i)

Also,3 – 3 = ( – )3 + 3. ( – ) 189 = (3)3 + 3. (3) [ . = 3 and 3 – 3 = 189] 189 = 27 + 9 . 189 – 27 = 9 . 162 = 9 .

162

9= .

. = 18Now, ( – )2 = ( + )2 – 4.

(3)2 = ( + )2 – 4 (18) [ – = 3 and . = 18] 9 = ( + )2 – 72 9 + 72 = ( + )2

( + )2 = 81Taking square root on both the sides we get,

+ = + 9 x2– (+) x + . = 0 [From (i)] x2 – (9)x + 18 = 0 or x2 – (– 9)x + 18 = 0 x2 – 9x + 18 = 0 or x2 + 9x + 18 = 0

The required quadratic equation is x2 – 9x + 18 = 0 or x2 + 9x + 18 = 0.


SCHOOL SECTION 125

26. If the difference of the roots of a quadratic equation is 4 and thedifference of their cubes is 208, find the quadratic equation. (4 marks)

Sol. Let and be the roots of a quadratic equation – = 4 and 3 – 3 = 208 [Given]

We know that,x2 – ( + ) x + . = 0 ......(i)

Also, 3 – 3 = ( – )3 + 3. (.) 208 = (4)3 + 3. (4) [.=4and3–3= 208] 208 = 64 + 12 . 208 – 64 = 12 . 144 = 12 .

144

12= .

. = 12Now, ( – )2 = ( + )2 – 4.

(4)2 = ( + )2 – 4 (12) [–= 4 and . = 12] 16 = ( + )2 – 48 16 + 48 = ( + )2

( + )2 = 64Taking square root on both the sides we get,

+ = + 8 x2 – ( + )x + . = 0 [From (i)] x2 – (8)x + 12 = 0 or x2 – (– 8)x + 12 = 0 x2 – 8x + 12 = 0 or x2 + 8x + 12 = 0

The required quadratic equation is x2 – 8x + 12 = 0 or x2 + 8x + 12 = 0

27. If one root of the quadratic equation ax2 + bx + c = 0 is the square of theother, show that b3 + a2c + ac2 = 3 abc. (4 marks)

To prove : b3 + a2c + ac2 = 3abcProof : ax2 + bx + c = 0 [Given]

Let and be the roots of given quadratic equation = 2 [Given]

+ =– b

a and. =

c

a

2. =c

a[ = 2]

3 =c

aWe know that,( + )3 = 3 + 3 + 3 . ( + )

– b

a

3

= (2)3 + 3 + 3 2.– b

a

– band

a

2

– b

a

3

3 = (3)2 + 3 + 33– b

a

– b

a

3

3 =c c c b

+ – 3 × ×a a a a

2 c

a

3

– b

a

3

3 =c c 3bc

+ –aa a

2

2 2

Multiplying throughout by a3 we get,– b3 = ac2 + a2c – 3abc

3abc =b3 + ac2 + a2c b3 + ac2 + a2c = 3abc

Hence proved.


SCHOOL SECTION126

28. if the sum of roots of the quadratic equation1 1

+x + p x + q =

1r

is zero.

Show that the product of the roots is –

2 2p + q2 . (5 marks)

Proof :1 1

+x + p x + q =

1

r[Given]

x + q + x + p

(x + p) (x + q) =1

r

2

2x + p + q

x + qx + px + pq =1

r r (2x + p + q) = x2 + qx + px + pq 2rx + rp + rq = x2 + qx + px + pq 0 = x2 + qx + px – 2rx + pq – pr – qr x2 + qx + px – 2rx + pq – pr – qr = 0 x2 + x (q + p – 2r) + pq – pr – qr = 0 x2 + (p + q – 2r) x + (pq – pr – qr) = 0 ......(i)

Comparing with ax2 + bx + c = 0 we have a = 1, b = p + q – 2r,c = pq – pr – qr

Let and be the roots of equation (i),

+ =– b

a =

– (p + q – 2r)

1 = – p – q + 2r

+ = 0 [Given] – p – q + 2r = 0 [ a + b = – p – q + 2r] 2r = p + q

r =p q

2

Also, . =c

a =

pq – pr – qr

1 = pq – pr – qr

. = pq – pr – qr

= pq – r (p + q)

= pq – p + q

2 (p + q)

p + qr =

2

=(p q)

pq –2

2

=2pq – (p + q)

2

2

=2pq – (p 2pq q )

2

2 2

=2pq – p – 2pq – q

2

2 2

=– p – q

2

2 2

. =– p – q

2

2 2

Hence proved.


SCHOOL SECTION 127

o Equations reducible to quadratic form :Sometimes given equation may not be a quadratic equation. But it can bereduced to quadratic equation by proper substitution. Then by solving it byany method, we get the values of new variable.

On resubstituting the values we get the value of the old variable.

EXERCISE - 2.9 (TEXT BOOK PAGE NO. 60) :Solve the following equations :

(i) x4 – 3x2 + 2 = 0 (3 marks)Sol. x4 – 3x2 + 2 = 0

(x2)2 – 3x2 + 2 = 0Substituting x2 = m we get,

m2 – 3m + 2 = 0 m2 – 2m – m + 2 = 0 m (m – 2) – 1 (m – 2) = 0 (m – 2) (m – 1) = 0 m – 2 = 0 or m – 1 = 0 m = 2 or m = 1

Resubstituting m = x2 we get,x2 = 2 or x2 = 1

Taking square roots throughout,

x = ± 2 or x = +1

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :29. Solve the following equations :(i) x4 – 29x2 + 100 = 0 (3 marks)Sol. x4 – 29x2 + 100 = 0


m2 – 29m + 100 = 0 m2 – 4m – 25m + 100 = 0 m(m – 4) – 25(m – 4) = 0 m – 4 = 0 or m – 25 = 0 m = 4 or m = 25


Taking square root on both the sides, we get,

x = ± 2 or x = ± 5

(ii) 7y4 – 25y2 + 12 = 0 (3 marks)Sol. 7y4 – 25y2 + 12 = 0

7(y2)2 – 25y2 + 12 = 0Substituting y2 = m we get,

7m2 – 25m + 12 = 0 7m2 – 21m – 4m + 12 = 0 7m(m – 3) – 4(m – 3) = 0 (m – 3)(7m – 4) = 0 m – 3 = 0 or 7m – 4 = 0

m = 3 or 7m = 4

m = 3 or m = 4

7


SCHOOL SECTION128

Resubstituting m = y2 we get,

y2 = 3 or y2 = 4

7Taking square root on both the sides, we get,

y = ± 3 or y = ± 2

7


(vii) 3x4 – 13x2 + 10 = 0 (3 marks)Sol. 3x4 – 13x2 + 10 = 0

3 (x2)2 – 13x2 + 10 = 0Substituting x2 = m we get,

3m2 – 13m + 10 = 0 3m2 – 3m – 10m + 10 = 0 3m (m – 1) – 10 (m – 1) = 0 (m – 1) (3m – 10) = 0 m – 1 = 0 or 3m – 10 = 0 m = 1 or 3m = 10

m = 1 or m = 10

3Resubstituting m = x2 we get,

x2 = 1 or x2 = 10


x = +1 or x = 10

±3

(v) x2 + 2

12x

= 7 (3 marks)

Sol. x2 + 12

x2 = 7

Multiplying throughout by x2 we get,x4 + 12 = 7x2


m2 – 7m + 12 = 0 m2 – 4m – 3m + 12 = 0 m (m – 4) – 3 (m – 4) = 0 (m – 4) (m – 3) = 0 m – 4 = 0 or m – 3 = 0 m = 4 or m = 3



x = + 2 or x = ± 3

(iv) 35y2 + 2

12y = 44 (3 marks)

Sol. 35y2 + 12

y2 = 44


SCHOOL SECTION 129

Multiplying throughout by y2 we get,35y4 + 12 = 44y2

35 (y2)2 – 44y2 + 12 = 0Substituting y2 = m we get,

35m2 – 44m + 12 = 0 35m2 – 14m – 30m + 12 = 0 7m (5m – 2) – 6 (5m – 2) = 0 (5m – 2) (7m – 6) = 0 5m – 2 = 0 or 7m – 6 = 0 5m = 2 or 7m = 6

m = 2

5or m =

6

7Resubstituting m = y2, we get;

y = 2

±5

or y = 6

±7

y = 2

5or y =

6

7

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :29. Solve the following equations :

(iii) 6m2 + 2

2m

= 7 (3 marks)

Sol. 6m2 +2

m2 = 7

Multiplying throughout by m2, we get6m4 + 2 = 7m2

6(m2)2 – 7m2 + 2 = 0Substituting m2 = x we get,

6x2 – 7x + 2 = 0 6x2 – 3x – 4x + 2 = 0 3x (2x – 1) – 2 (2x – 1) = 0 (2x – 1)(3x – 2) = 0 2x – 1 = 0 or 3x – 2 = 0 2x = 1 or 3x = 2

x = 1

2or x =

2

3Resubstituting x = m2 we get,

m2 = 1

2or m2 =

2


m = ± 1

2or m = ±

2

3


(viii) 2y2 + 2

15y = 12 (3 marks)

Sol. 2y2 + 15

y2 = 12


SCHOOL SECTION130

Multiplying throughout by y2 we get,2y4 + 15 = 12y2

2 (y2)2 – 12y2 + 15 = 0Substituting y2 = m we get,

2m2 – 12m + 15 = 0Comparing with am2 + bm + c = 0 we a = 2, b = – 12, c = 15

b2 – 4ac = (– 12)2 – 4 (2) (15)= 144 – 120= 24

m =– b ± b – 4ac

2a

2

=( )– –12 ± 24

2(2)

=( )– –12 ± 24

2(2)

=12 ± 2 6

4

=( )2 6 ± 6

4

=6 ± 6

2

m = 6 + 6

2or m =

6 – 6

2Resubstituting m = y2 we get,

y2 = 6 + 6

2or y2 =

6 – 6


y = 6 + 6

±2

or y = 6 – 6

±2

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :29. Solve the following equations :(x) 2 (y2 – 6y)2 – 8 (y2 – 6y + 3) – 40 = 0 (3 marks)Sol. 2 (y2 – 6y)2 – 8 (y2 – 6y + 3) – 40 = 0

Substituting y2 – 6y = m we get, 2m2 – 8 (m + 3) – 40 = 0 2m2 – 8m – 24 – 40 = 0 2m2 – 8m – 64 = 0

Dividing throughout by 2 we get,m2 – 4m – 32 = 0

m2 – 8m + 4m – 32 = 0 m (m – 8) + 4 (m – 8) = 0 (m – 8) (m + 4) = 0 m – 8 = 0 or m + 4 = 0 m = 8 or m = – 4

Resubstituting m = y2 – 6y we get,y2 – 6y = 8 .....(i) or y2 – 6y = – 4 ........(ii)

From (i), y2 – 6y = 8 y2 – 6y – 8 = 0


SCHOOL SECTION 131

Comparing with ay2 + by + c = 0 we have a = 1, b = – 6, c = – 8b2 – 4ac = (– 6)2 – 4 (1) (– 8)

= 36 + 32= 68

y =– b ± b – ac

2a

2

=– (–6) ± 68

2 (1)

=6 ± 4 ×17

2

=6 ± 2 17

2

= 2 3 ± 17

2

= 3 ± 17

y = 3 + 17 or y = 3 – 17From (ii), y2 – 6y = – 4

y2 – 6y + 4 = 0Comparing with ay2 + by + c = 0 we have a = 1, b = – 6, c = 4

b2 – 4ac = (– 6)2 – 4 (1) (4)= 36 – 16= 20

y =– b ± b – 4ac

2a

2

=– (–6) ± 20

2 (1)

=6 ± 4 × 5

2

=6 ± 2 5

2

= 2 3 ± 5

2= 3 ± 5

y = 3 + 5 or 3 – 5

y = 3 + 17 or y = 3 – 17 or y = 3 + 5 or y = 3 – 5

(ix) (y2 + 5y) (y2 + 5y – 2) – 24 = 0 (3 marks)Sol. (y2 + 5y) (y2 + 5y – 2) – 24 = 0

Substituting y2 + 5y = m we get, m (m – 2) – 24 = 0 m2 – 2m – 24 = 0 m2 – 6m + 4m – 24 = 0 m (m – 6) + 4 (m – 6) = 0 (m – 6) (m + 4) = 0 m – 6 = 0 or m + 4 = 0 m = 6 or m = – 4


SCHOOL SECTION132

Resubstituting m = y2 + 5y we get,y2 + 5y = 6 ......(i) or y2 + 5y = – 4 .....(ii)

From (i), y2 + 5y = 6 y2 + 5y – 6 = 0 y2 – y + 6y – 6 = 0 y (y – 1) + 6 (y – 1) = 0 (y – 1) (y + 6) = 0 y – 1 = 0 or y + 6 = 0 y = 1 or y = – 6

From (ii), y2 + 5y = – 4 y2 + 5y + 4 = 0 y2 + 4y + y + 4 = 0 y (y + 4) + 1 (y + 4) = 0 (y + 4) (y + 1) = 0 y + 4 = 0 or y + 1 = 0 y = – 4 or y = – 1

y = 1 or y = – 6 or y = – 4 or y = – 1.

(iv) (p2 + p)(p2 + p – 3) = 28 (4 marks)Sol. (p2 + p)(p2 + p – 3) = 28

Substituting p2 + p = m we get, m (m – 3) = 28 m2 – 3m – 28 = 0 m2 – 7m + 4m – 28 = 0 m (m – 7) + 4 (m – 7) = 0 (m – 7) (m + 4) = 0 m – 7 = 0 or m + 4 = 0 m = 7 or m= –4

Resubstituting m = p2 + p we get,p2 + p = 7 or p2 + p = – 4

p2 + p – 7= 0...........(i) or p2 + p + 4 = 0 ........(ii)From (i), p2 + p – 7 = 0Comparing with ap2 + bp + c = 0 we have a =1, b = 1, c = – 7

b2 – 4ac = (1)2 – 4 (1) (–7)= 1 + 28= 29

p =– b ± b – 4ac

2a

2

=–1 ± 29

2(1)

=–1 ± 29

2From (ii), p2 + p + 4 = 0Comparing with ap2 + bp + c = 0 we have a = 1, b =1, c = 4

b2 – 4ac = (1)2 – 4 (1)(4)= 1 – 16= – 15

b2 – 4ac < 0


SCHOOL SECTION 133

The roots of this equation are not real hence not considered.

p =–1 ± 29

2

p = –1 29

2

or p =

–1 – 29

2


Solve the following equations :(ii) (x2 + 2x) (x2 + 2x – 11) + 24 = 0 (4 marks)Sol. (x2 + 2x) (x2 + 2x – 11) + 24 = 0

Substituting x2 + 2x = m we get,m (m – 11) + 24 = 0


Resubstituting m = x2 + 2x we get,x2 + 2x = 8 or x2 + 2x = 3

x2 + 2x – 8 = 0 ....... (i) or x2 + 2x – 3 = 0 ......(ii)From (i), x2 + 2x – 8 = 0

x2 + 4x – 2x – 8 = 0 x(x + 4) – 2 (x + 4) = 0 (x + 4) (x – 2) = 0 x + 4 = 0 or x – 2 = 0 x = – 4 or x = 2

From (ii), x2 + 2x – 3 = 0 x2 + 3x – x – 3 = 0 x (x + 3) – 1(x + 3) = 0 (x + 3) (x – 1) = 0 x + 3 = 0 or x – 1 = 0 x = – 3 or x = 1

x = – 4 or x = 2 or x = – 3 or x = 1.

(vi) (x2 + x) (x2 + x – 7) + 10 = 0 (4 marks)Sol. (x2 + x) (x2 + x – 7) + 10 = 0

Substituting x2 + x = m we get,m (m – 7) + 10 = 0


Resubstituting m = x2 + x we get,


SCHOOL SECTION134

x2 + x = 5 or x2 + x = 2 x2 + x – 5 = 0 ....... (i) or x2 + x – 2 = 0 ......(ii)

From (i), x2 + x – 5 = 0Comparing with ax2 + bx + c = 0 we have a = 1, b = 1, c = – 5

b2 – 4ac = (1)2 – 4(1) (– 5)= 1 + 20= 21

x =– b b – 4ac

2a

2

x =–1 21

2 × 1

x =–1 21

2

x = –1 21

2

or x =

–1 – 21

2From (ii), x2 + x – 2 = 0

x2 + 2x – 1x – 2 = 0 x (x + 2) – 1(x + 2) = 0 (x + 2) (x – 1) = 0 x + 2 = 0 or x – 1 = 0 x = – 2 or x = 1

x = –1 21

2

or x =

–1 – 21

2 or x = – 2 or x = 1.


(iii)

-22

1 12 x + 9 x+ +14=0

x x (5 marks)

Sol.1 1

2 x + – 9 x + + 14 = 0x x

22 .......(i)

Substituting 1

x +x

= m

Squaring both the sides we get,

1x +

x

2

= m2

(x)2 + 2 × x × 1

x +

1

x

2

= m2

x2 + 2 + 1

x2 = m2

x2 + 1

x2 = m2 – 2

Equation (i) becomes 2 (m2 – 2) – 9m + 14 = 0 2m2 – 4 – 9m + 14 = 0 2m2 – 9m + 10 = 0 2m2 4m – 5m + 10 = 0


SCHOOL SECTION 135

2m (m – 2) – 5 (m – 2) = 0 (m – 2) (2m – 5) = 0 m – 2 = 0 or 2m – 5 = 0 m = 2 or 2m = 5

m = 2 or m = 5

2

Resubstituting m = 1

xx

we get,

x + 1

x = 2 ..... (ii) or x +

1

x =

5

2 ..... (iii)

From (ii), x + 1

x= 2

Multiplying throughout by x we get,x2 + 1 = 2x

x2 – 2x + 1 = 0 (x – 1)2 = 0

Taking square root on both the sides we get,x – 1 = 0

x = 1

From (iii), x + 1

x=

5

2Multiplying throughout by 2x, we get;

2x2 + 2 = 5x 2x2 – 5x + 2 = 0 2x2 – 4x – x + 2 = 0 2x (x – 2) – 1 (x – 2) = 0 (x – 2) (2x – 1) = 0 x – 2 = 0 or 2x – 1 = 0 x = 2 or 2x = 1

x = 2 or x = 1

2

x = 1 or x = 2 or x = 1

2

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 166) :29. Solve the following equations :

(viii)

22

1 112 x + – 56 x + + 89 = 0

xx(5 marks)

Sol.1 1

12 x + – 56 x + + 89 = 0xx

22 .....(i)

Substituting 1

x +x

= m


1x +

x

2

= m2

x2 + 2 × x × 1

x +

1

x2 = m2

x2 + 2 + 1

x2 = m2

x2 + 1

x2 = m2 – 2


SCHOOL SECTION136

Equation (i) becomes,12 (m2 – 2) – 56m + 89 = 0

12m2 – 24 – 56m + 89 = 0 12m2 – 56m + 65 = 0 12m2 – 30m – 26m + 65 = 0 6m (2m – 5) – 13 (2m – 5) = 0 (2m – 5) (6m – 13) = 0 2m – 5 = 0 or 6m – 13 = 0 2m = 5 or 6m = 13

m = 5

2or m =

13

6


xx

we get,

x + 1

x =

5

2.....(ii) or x +

1

x =

13

6 .....(iii)

From (ii),1

xx

=5

2Multiplying throughout by 2x we get,

2x2 + 2 = 5x 2x2 – 5x + 2 = 0 2x2 – 4x – x + 2 = 0 2x (x – 2) – 1 (x – 2) = 0 (x – 2) (2x – 1) = 0 x – 2 = 0 or 2x – 1 = 0 x = 2 or 2x = 1

x = 2 or x = 1

2

From (iii), x + 1

x=

13


6x2 + 6 = 13x 6x2 – 13x + 6 = 0 6x2 – 9x – 4x + 6 = 0 3x (2x – 3) – 2 (2x – 3) = 0 (2x – 3) (3x – 2) = 0 2x – 3 = 0 or 3x – 2 = 0 2x = 3 or 3x = 2

x = 3

2or x =

2

3

x = 2 or x = 1

2 or x =

3

2 or x =

2

3

(v) 3

22

1x +

x – 4

1x –

x – 6 = 0 (5 marks)

Sol. 31

x +x

22 – 4

1x –

x

– 6 = 0 ......(i)

Substituting 1

x –x

= m

squaring both the sides we get,

1x –

x

2

= m2

x2 – 2 . x . 1

x +

1

x

2

= m2


SCHOOL SECTION 137

x2 – 2 + 1

x2 = m2

x2 + 1

x2 = m2 + 2

Equation (i) becomes, 3 (m2 + 2) – 4m – 6 = 0 3m2 + 6 – 4m – 6 = 0 3m2 – 4m = 0

m (3m – 4) = 0 m = 0 or 3m - 4 = 0 m = 0 or 3m = 4

m = 0 or m = 4

3


x –x

we get,

1x –

x = 0 ........(ii) or

1x –

x =

4

3 .......(iii)

From (ii),1

x –x

= 0

Multiplying throughout by x, we get;x2 – 1 = 0

x2 – 12 = 0 (x + 1)(x – 1) = 0 x + 1 = 0 or x – 1 = 0 x = – 1 or x = 1

From (iii),1

x –x

=4


3x2 – 3 = 4x 3x2 – 4x – 3 = 0

Comparing with ax2 + bx + c = 0 we have a =3, b = – 4, c = – 3b2 – 4ac = (– 4)2 – 4 (3)(–3)

= 16 + 36= 52

x =– b + b – 4ac

2a

2

=– (– 4) ± 52

2(3)

=4 ± 4×13

6

=4 ± 2 13

6

=( )2 2 ± 13

6

=2 ± 13

3

x = 2 + 13

3or x =

2 – 13

3

x = – 1 or x = 1 or x = 2 + 13

3 or x =

2 – 13

3


SCHOOL SECTION138

(vi) 9

22

1x +

x – 3

1x –

x – 20 = 0 (5 marks)

Sol. 9 1

x +x

22 – 3

1x –

x

– 20 = 0 ..........(i)

Substituting 1

x –x

= m


1x –

x

2

= m2

x2 – 2 + 1

x2 = m2

x2 + 1

x2 = m2 + 2

Equation (i) becomes,9(m2 + 2) – 3m – 20 = 0

9m2 + 18 – 3m – 20 = 0 9m2 – 3m – 2 = 0 9m2 + 3m – 6m – 2 = 0 3m (3m + 1) – 2 (3m + 1) = 0 (3m + 1) (3m – 2) = 0 3m + 1= 0 or 3m – 2 = 0 3m = – 1 or 3m = 2

m = 1

–3

or m = 2

3


x –x

we get,

1x –

x=

1–

3 ..........(ii) or

1x –

x=

2

3 ...........(iii)

From (ii),1

x –x

=1

–3

Multiplying throughout by 3x we get, 3x2 – 3 = – x 3x2 + x – 3 = 0


= 1 + 36= 37

x =– b ± b – 4ac

2a

2

=–1 ± 37

2(3)

=–1 ± 37

6

x = 1 + 37

6or x =

–1 – 37

6

From (iii),1

x –x

=2


3x2 – 3 = 2x 3x2 – 2x – 3 = 0


SCHOOL SECTION 139

Comparing with ax2 + bx + c = 0 we have a = 3, b = –2, c = – 3b2 – 4ac = (–2)2 – 4 (3)(– 3)

= 4 + 36= 40

x =– b ± b – 4ac

2a

2

=– 2 ± 40

2(3)

=– 2 ± 4 ×10

6

=– 2 ± 2 10

6

=– 1 ± 10

3

x = – 1 + 10

3or

– 1 – 10

3

x = 1 + 37

6 or x =

–1 – 37

6 or x =

– 1 + 10

3 or x =

– 1 – 10

3

(vii) 30

22

1x +

x – 77

1x –

x – 12 = 0 (5 marks)

Sol. 30 1

x +x

22 – 77

1x –

x

– 12 = 0 .........(i)

Substituting 1

x –x

= m


1x –

x

2

= m2

x2 – 2 × x × 1

x +

1

x2 = m2

x2 – 2 + 1

x2 = m2

x2 +1

x2 = m2 + 2

Equation (i) becomes,30 (m2 + 2) – 77m – 12 = 0

30m2 + 60 – 77m – 12 = 0 30m2 – 77m + 48 = 0 30m2 – 45m – 32m + 48 = 0 15m (2m – 3) – 16 (2m – 3) = 0 (2m – 3)(15m – 16) = 0 2m = 3 or 15m = 16

m = 3

2or m =

16

15

Resubstituting the m =1

x –x

we get,

1x –

x =

3

2 ........(ii) or

1x –

x =

16

15 ......(iii)


SCHOOL SECTION140

From (ii),1

x –x

=3


2x2 – 2 = 3x 2x2 – 3x – 2 = 0 2x2 – 4x + x – 2 = 0 2x (x – 2)+ 1 (x – 2) = 0 (x – 2)(2x + 1) = 0 x – 2 = 0 or 2x + 1 = 0 x = 2 or 2x = – 1

x = 2 or x = –1

2

From (iii),1

x –x

=16


15x2 – 15 = 16x 15x2 – 16x – 15 = 0 15x2 – 25x + 9x – 15 = 0 5x (3x – 5) + 3 (3x – 5) = 0 (3x – 5)(5x + 3) = 0 3x – 5 = 0 or 5x + 3 =0 3x = 5 or 5x = – 3

x = 5

3or x =

– 3

5

x = 2 or x = –1

2 or x =

5

3 or x =

– 3

5

WORD PROBLEMS

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 164) :2. From an equations for the following examples.(i) The sum of a natural number ‘x’ and its square is 30. (1 mark)Sol. A natural number is ‘x’

Its square is x2

As per the given condition,x + x2 = 30

x2 + x - 30 = 0

(ii) The product of two numbers ‘y’ and y – 3 is 42. (1 mark)Sol. Two numbers are ‘y’ and y – 3.

As per the given condition,y (y – 3) = 42

y2 – 3y – 42 = 0

(iii) The sum of a natural number ‘x’ and its reciprocal is 376

. (1 mark)

Sol. A natural number is ‘x’.

It’s reciprocal is 1

x.

As per the given condition,

x + 1

x=

37


6x2 + 6 = 37x 6x2 – 37x + 6 = 0


SCHOOL SECTION 141

(iv) The digit at ten’s place of a two digit number is greater than the squareof digit at unit’s place(x) by 5 and the number formed is 61. (1 mark)

Sol. The digit is unit’s place of a two digit number is ‘x’. The digit in ten’s place is x2 + 5

As per the given condition,10 (x2 + 5) + 1 (x) = 61

10x2 + 50 + x – 61 = 0 10x2 + x – 11 = 0

(v) The length of a rectangle (x) is greater than its breadth by 3 cm. Thearea of a rectangle is 70 sq.cm (1 mark)

Sol. The length of a rectangle is ‘x’ cm. It’s breadth is (x – 3) cm.

As per the given condition, Area of rectangle = Length × Breadth 70 = x × (x – 3) 70 = x2 – 3x 0 = x2 – 3x – 70 x2 – 3x – 70 = 0


6. The sum of a natural number and its reciprocal is 103

. Find the number.

(3 marks)Sol. Let natural number be ‘x’

Its reciprocal is 1

xFrom the given condition,

x + 1

x =

10

3Multiplying throughout by 3x,3x2 + 3 = 10x

3x2 – 10x + 3 = 0 3x2 – 9x – x + 3 = 0 3x (x – 3) – 1 (x – 3) = 0 (3x – 1) (x – 3) = 0 3x – 1 = 0 or x – 3 = 0 3x = 1 or x = 3

x = 1

3or x = 3

x 1

3 because x is natural number

x = 3

The natural number is 3.

8. Three times the square of a natural numbers is 363. Find the numbers.(3 marks)

Sol. Let the natural number be ‘x’From the given condition,3x2 = 363

x2 = 363

3


SCHOOL SECTION142

x2 = 121 x = + 11 [Taking square roots]

x – 11 because x is a natural number x = 11


11. The sum ‘S’ of the first ‘n’ natural numbers is given by S = n (n + 1)

2.

Find ‘n’, if the sum (S) is 276. (3 marks)

Sol. S = n (n + 1)

2From the given condition,

276 = n (n + 1)

2 552 = n2 + n n2 + n – 552 = 0 n2 + 24n – 23n – 552 = 0 n (n + 24) (n – 23) = 0 n + 24 = 0 or n – 23 = 0 n = – 24 or n = 23

n – 24 because ‘n’ cannot be negative

n = 23

WORD PROBLEMS BASED ON NUMBERS

EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :1. The sum of the squares of two consecutive natural numbers is 113.

Find the numbers. (3 marks)Sol. Let the two consecutive natural numbers be x and x + 1

As per the given condition,x2 + (x + 1)2 = 113

x2 + x2 + 2x + 1 = 113 2x2 + 2x + 1 – 113 = 0 2x2 + 2x – 112 = 0

Dividing throughout by 2 we get,x2 + x – 56 = 0

x2 + 8x – 7x – 56 = 0 x (x + 8) – 7 (x + 8) = 0 (x + 8) (x – 7) = 0 x + 8 = 0 or x – 7 = 0 x = – 8 or x = 7 x is a natural number x -8

Hence, x = 7And x + 1 = 7 + 1 = 8

The two consecutive natural numbers are 7 and 8 respectively.

4. The sum of the squares of two consecutive even natural numbers is100. Find the numbers. (3 marks)

Sol. Let the two consecutive even natural numbers be x and x +2.As per the given condition.

x2 + (x + 2)2 = 100 x2 + x2 + 4x + 4 = 100


SCHOOL SECTION 143

2x2 + 4x + 4 – 100 = 0 2x2 + 4x – 96 = 0

Dividing throughout by 2 we get,x2 + 2x – 48 = 0

x2 – 6x + 8x – 48 = 0 x (x – 6) + 8 (x – 6) = 0 (x – 6) (x + 8) = 0 x – 6 = 0 or x + 8 = 0 x = 6 or x = – 8

x is an natural number x – 8

Hence x = 6And x + 2 = 6 + 2 = 8

The two consecutive even natural numbers are 6 and 8 respectively.

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :30. Solve the following :(i) Three consecutive odd natural numbers are such that the product of

the first and third is greater than four times the middle by 1. Find thenumbers. (3 marks)

Sol. Let the third consecutive odd natural number be x, x + 2 and x + 4As per the given condition,

x (x + 4) = 4 (x + 2) + 1 x2 + 4x = 4x + 8 + 1 x2 = 9

Taking square root on both the sides we get,x = + 3

x is a natural number x – 3Hence, x = 3x + 2 = 3 + 2 = 5 and x + 4 = 3 + 4 = 7

The 3 consecutive odd natural numbers are 3, 5 and 7 respectively.

(xiii) The sum of the squares of five consecutive natural numbers is 1455.find them. (4 marks)

Sol. Let the five consecutive natural numbers be x, x + 1, x + 2, x + 3 andx + 4 respectively.As per the given condition,x2 + (x + 1)2 + (x + 2)2 + (x + 3)2 + (x + 4)2 = 1455

x2 + x2 + 2x + 1 + x2 + 4x + 4 + x2 + 6x +9 + x2 + 8x + 16 – 1455 = 0 5x2 + 20x + 30 – 1455 = 0 5x2 + 20x – 1425 = 0


x2 – 15x + 19x – 285 = 0 x (x – 15) + 19 (x – 15) = 0 (x – 15) (x + 19) = 0 x – 15 = 0 or x + 19 = 0 x = 15 or x = – 19 x is a natural number x – 19

Hence x = 15 x + 1 = 15 + 1 = 16 x + 2 = 15 + 2 = 17 x + 3 = 15 + 3 = 18 x + 4 = 15 + 4 = 19

The required five consecutive natural numbers are 15, 16, 17, 18and 19 respectively.


SCHOOL SECTION144

(x) The product of four consecutive positive integers is 840. find the largestnumber. (5 marks)

Sol. Let the four consecutive positive integers be x, x + 1, x + 2 and x + 3As per the given condition,x × (x + 1) × (x + 2) × (x + 3) = 840

x (x + 3) × (x + 1) (x + 2) = 840 (x2 + 3x) × (x2 + 2x + x + 2) = 840 (x2 + 3x) (x2 + 3x + 2) = 840

Substituting x2 + 3x = m we get,m (m + 2) = 840

m2 + 2m – 840 = 0 m2 + 30m – 28m – 840 = 0 m (m + 30) – 28 (m + 30) = 0 (m + 30) (m – 28) = 0 m + 30 = 0 or m – 28 = 0 m = – 30 or m = 28

Resubstituting m = x2 + 3x we get,x2 + 3x = – 30 ......(i) or x2 + 3x = 28 ......(ii)

From (i), x2 + 3x = – 30 x2 + 3x + 30 = 0

Comparing with ax2 + bx + c = 0 we have a = 1, b = 3, c = 30b2 – 4ac = (3)2 – 4 (1) (30)

= 9 – 120= – 111

b2 – 4ac < 0 The roots of the above quadratic equation are not real.

Hence not considered.From (ii), x2 + 3x = 28

x2 + 3x – 28 = 0 x2 + 7x – 4x – 28 = 0 x (x + 7) – 4 (x + 7) = 0 (x + 7) (x – 4) = 0 x + 7 = 0 or x – 4 = 0 x = – 7 or x = 4 x is positive integer x – 7

Hence x = 4 x + 3 = 4 + 3 = 7

The largest required number is 7.

(iv) Divide 40 into two parts such that the sum of their reciprocals is 875

.

(4 marks)Sol. Sum of two parts is 40

Let one of the part is x The other part is 40 – x

As per the given condition,1 1

x 40 – x =

8

75

40 – x + x

x (40 – x) =8

75

40

40x – x2 =8

75 40 (75) = 8 (40x – x2)


SCHOOL SECTION 145

40 × 75

8= 40x – x2

375 = 40x – x2

x2 – 40x + 375 = 0 x2 – 25x – 15x + 375 = 0 x (x – 25) – 15 (x – 25) = 0 (x – 25) (x – 15) = 0 x – 25 = 0 or x – 15 = 0 x = 25 or x = 15

If x = 25 or if x = 15then 40 – x = 40 – 25 = 15 then 40 – x = 40 – 15 = 25

The two parts are 25 and 15.

(xvi) The difference between two positive integers is 2 and the differencebetween their cubes is 56. Find the numbers. (3 marks)

Sol. Let the smaller positive integer be x The bigger positive integer is x + 2

As per the given condition,(x + 2)3 – x3 = 56

x3 + 3 × (x2) × (2) + 3 × (x) × 22 + 23 – x3 – 56 = 0 x3 + 6x2 + 12x + 8 – x3 – 56 = 0 6x2 + 12x – 48 = 0


x2 + 4x – 2x – 8 = 0 x (x + 4) – 2 (x + 4) = 0 (x + 4) (x – 2) = 0 x + 4 = 0 or x – 2 = 0 x = – 4 or x = 2 x is a positive integer x – 4

Hence x = 2 x + 2 = 2 + 2 = 4

The two positive integers are 2 and 4.

(vi) The sum of four times a number and three times its reciprocal is 7. Findthat number. (3 marks)

Sol. Let the number be x

It’s reciprocal is 1

xAs per the given condition,

4 (x) + 3 1

x

= 7

4x + 3

x= 7

Multiplying throughout by x we get,4x2 + 3 = 7x

4x2 – 7x + 3 = 0 4x2 – 4x – 3x + 3 = 0 4x (x – 1) – 3 (x – 1) = 0 (x – 1) (4x – 3) = 0


SCHOOL SECTION146

x – 1 = 0 or 4x – 3 = 0 x = 1 or 4x = 3

x = 1 or x = 3

4

The number is 1 or 3

4.

(viii) A natural number is greater then three times its square root by 4. Findthe number. (3 marks)

Sol. Let the square root of natural number be x The natural number = x2

As per the given condition,x2 = 3x + 4

x2 – 3x – 4 = 0 x2 – 4x + 1x – 4 = 0 (x – 4) (x + 1) = 0 x – 4 = 0 or x + 1 = 0 x = 4 or x = – 1

Square root of natural number cannot be negative x – 1 x = 4 x2 = 42 x2 = 16


EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :5. A natural number is greater than twice its square root by 3. Find the

number. (3 marks)Sol. Let square root of the natural number be x

The natural number = x2

As per the given condition,x2 = 2x + 3

x2 – 2x – 3 = 0 x2 – 3x + 1x – 3 = 0 x (x – 3) + 1 (x – 3) = 0 (x – 3) (x + 1) = 0 x – 3 = 0 or x + 1 = 0 x = 3 or x = – 1 Square root of a natural number cannot be negative x – 1 x = 3 x2 = 9


10. A natural number is greater than the other by 5. The sum of theirsquares is 73. Find those numbers. (3 marks)

Sol. Let the other natural number be x. The first natural number is x + 5.


x2 + x2 + 10x + 25 – 73 = 0 2x2 + 10x – 48 = 0


SCHOOL SECTION 147


x2 – 3x + 8x – 24 = 0 x (x – 3) + 8 (x – 3) = 0 (x – 3) (x + 8) = 0 x – 3 = 0 or x + 8 = 0 x = 3 or x = -8 The number is a natural number x – 8

Hence x = 3And x + 5 = 3 + 5 = 8

The natural numbers are 3 and 8.

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 168) :30. Solve the following :(xix) The divisor and quotient of the number 6123 are same and the remainder

is half the divisor. Find the divisor. (4 marks)Sol. Let divisor of 6123 be ‘x’

Divisor = Quotient [Given] Quotient = x

Remainder = x

2[Given]

We know,Dividend = Divisor × Quotient + Remainder

6123 = x . x + x

2Multiplying throughout by 2,

2 (6123) = 2x2 + x 12246 = 2x2 + x 2x2 + x – 12246 = 0 2x2 + 157x – 156x – 12246 = 0 x (2x + 157) – 78 (2x + 157) = 0 (2x + 157) (x – 78) = 0 2x + 157 = 0 or (x – 78) = 0 2x + 157 = 0 or x – 78 = 0

x = – 157

2or x = 78

x = – 157

2 cannot be acceptable because divisor cannot be negative.

x = 78

The divisor of 6123 is 78.

WORD PROBLEMS BASED ON AGE

EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :2. Tinu is younger than Pinky by three years. The product of their ages is

180. Find their ages. (3 marks)Sol. Tinu’s age be ‘x’ years

Pinky’s age is (x + 3) yearsAs per the given condition,

x (x + 3) = 180 x2 + 3x – 180 = 0 x2 – 12x + 15x – 180 = 0


SCHOOL SECTION148

x (x – 12) + 15 (x – 12) = 0 (x – 12) (x + 15) = 0 x – 12 = 0 or x + 15 = 0 x = 12 or x = – 15 Age cannot be negative. x – 15 x = 12

And x + 3 = 12 + 3 = 15

Tinu’s age is 12 years and Pinky’s is 15 years.

7. The sum of the ages of father and his son is 42 years. The product oftheir ages is 185, find their ages. (3 marks)

Sol. Sum of ages of father and son = 42 yearsLet father’s age be x years

Son’s age = (42 – x) years The age of his son is (42 – x) years.

As per the given condition,x (42 – x) = 185

42x – x2 = 185 0 = x2 – 42x + 185 x2 – 42x + 185 = 0 x2 – 5x – 37x + 185 = 0 x (x – 5) – 37 (x – 5) = 0 (x – 5) (x – 37) = 0 x – 5 = 0 or x – 37 = 0 x = 5 or x = 37

If x = 5, father’s age is less than son’s age. x 5

Hence x = 37And 42 – x = 42 – 37 = 5

The father’s age is 37 years and son’s age 5 years.

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :30. Solve the following :

(xv) Two years ago my age was 412

times the age of my son. Six years ago,

my age was twice the square of the age of my son. What is the presentage of my son ? (4 marks)

Sol. Let the present age of my son be x yearsTwo years ago,My son’s age = (x – 2) years

My age = 1

42

(x – 2) years = 9

2 (x – 2) years

Six years ago,My son’s age = (x – 6) yearsMy age = 2 (x – 6)2 yearsAs per the given condition,

9

2 (x – 2) – 2 (x – 6)2 = 4

Multiplying throughout by 2 we get,9 (x – 2) – 4 (x – 6)2 = 8

9x – 18 – 4 (x2 – 12x + 36) = 8


SCHOOL SECTION 149

9x – 18 – 4x2 + 48x – 144 = 8 – 4x2 + 57x – 162 = 8 0 = 4x2 – 57x + 162 + 8 4x2 – 57x + 162 + 8 = 0 4x2 – 57x + 170 = 0 4x2 – 40x – 17x + 170 = 0 4x (x – 10) – 17 (x – 10) = 0 (x – 10) (4x – 17) = 0 x – 10 = 0 or 4x – 17 = 0 x = 10 or 4x = 17

x = 10 or x = 17

4

The age of son six years ago becomes negative for x = 17

4

x 17

4Hence x = 10

The present age of son is 10 years.

WORD PROBLEMS BASED ON GEOMETRIC FIGURES

EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :3. The length of the rectangle is greater than its breadth by 2 cm. The area

of the rectangle is 24 sq.cm, find its length and breadth. (3 marks)Sol. Let the breadth of the rectangle be ‘x’ cm.

It’s length is (x + 2) cm.As per the given condition.

Area of rectangle = Length × Breadth 24 = (x + 2) × x 24 = x2 + 2x 0 = x2 + 2x – 24 x2 + 2x – 24 = 0 x2 – 4x + 6x – 24 = 0 x (x – 4) + 6(x – 4) = 0 (x – 4) (x + 6) = 0 x – 4 = 0 or x + 6 = 0 x = 4 or x = -6 The breadth of rectangle cannot be negative. x -6

Hence x = 4 and x + 2 = 4 + 2 = 6

The length of rectangle is 6 cm and its breadth is 4 cm.

12. A rectangular playground is 420 sq.m. If its length is increases by 7 mand breadth is decreased by 5 metres, the area remains the same. Findthe length and breadth of the playground ? (5 marks)

Sol. Let the length of a rectangular playground be ‘x’ m. The area of playground is 420 sq.m.

It’s breadth is 420

xm

New length = (x + 7) m


SCHOOL SECTION150

New breadth = 420

– 5 mx

As per the given condition, Area of new rectangle = Length × Breadth

New area = (x + 7) 420

– 5x

420 = (x + 7) × 420

5x

420 = 420 – 5x + 2940

x – 35

Multiplying throughout by x, we get;0 = – 5x2 + 2940 – 35x

5x2 + 35x – 2940 = 0 x2 + 7x – 588 = 0 [Dividing throughout by 5] x2 – 21x + 28x – 588 = 0 x (x – 21) + 28 (x – 21) = 0 (x – 21) (x + 28) = 0 x – 21 = 0 or x + 28 = 0 x = 21 or x = -28 The length of playground cannot be negative. x -28

Hence x = 21

And 420

x =

420

21 = 20

The length of a rectangular playground is 21 m and its breadth is 20 m.

9. The length of one diagonal of a rhombus is less than the second diagonalby 4 cm. The area of the rhombus is 30 sq.cm. Find the length of thediagonals. (4 marks)

Sol. Let the length of other diagonal of a rhombus be ‘x’ cm. The length of first diagonal is (x + 4) cm.

Area of rhombus = 1

2 × Product of length of diagonals

Area of rhombus = 1

2 × x × (x + 4)

As per the given condition,1

2x (x + 4) = 30

x (x + 4) = 60 x2 + 4x – 60 = 0 x2 + 10x – 6x – 60 = 0 x (x + 10) – 6 (x + 10) = 0 (x + 10) (x – 6) = 0 x + 10 = 0 or x – 6 = 0 x = – 10 or x = 6 The length of diagonal of the rhombus cannot be negative. x – 10

Hence x = 6And x + 4 = 6 + 4 = 10

The length of smaller diagonal of a rhombus is 6 cm and bigger diagonalis 10 cm.


SCHOOL SECTION 151

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :30. Solve the following :(iii) One diagonal of a rhombus is greater than other by 4 cm. If the area of

the rhombus is 96 cm2, find the side of the rhombus. (5 marks)Sol. Let the length of smaller diagonal of a rhombus be x cm

The length of bigger diagonal is (x + 4) cmAs per the given condition,

Area of rhombus =1

2 × Diagonal 1 × Diagonal 2


1

2 × x × (x + 4) = 96

x2 + 4x = 192 x2 + 4x – 192 = 0 x2 – 12x + 16x – 192 = 0 x (x – 12) + 16 (x – 12) = 0 (x – 12) (x + 16) = 0 x – 12 = 0 or x + 16 = 0 x = 12 or x = – 16 The length of diagonal of the rhombus cannot be negative x – 16

Hence x = 12 x + 4 = 12 + 4 = 16

Considering ABCD a rhombus l (AC) = 12 cm l (BD) = 16 cm

O is the intersection point of diagonal AC and BD

l (AO) = 1

2l (AC) =

1

2 × 12 = 6 cm

l (BO) = 1

2l (BD) =

1

2 × 16 = 8 cm

AOB is a right angled triangleIn right angled AOB,[l (AB)]2 = [l (AO)]2 + [l (BO)]2 [By Pythagoras theorem]

[l (AB)]2 = (6)2 + (8)2

[l (AB)]2 = 36 + 64 [l (AB)]2 = 100

Taking square root on both the sides we get,l (AB) = 10 cm

The side of a rhombus is 10 cm.


30. Solve the following :(ix) The sum of the areas of two squares is 400sq.m. If the difference between

their perimeters is 16 m, find the sides of two square. (4 marks)Sol. Difference between the perimeters of two squares is 16 m

The difference between the sides of the same squares is 4 mLet the side of smaller square be x cm

The side of bigger square is (x + 4) cm

Diagonal of a rhombus areperpendicular bisectors ofeach other

}

A D

B C

O


SCHOOL SECTION152

Area of square = (side)2


x2 + x2 + 8x + 16 – 400 = 0 2x2 + 8x – 384 = 0


x2 – 12x + 16x – 192 = 0 x (x – 12) + 16 (x – 12) = 0 (x – 12) (x + 16) = 0 x – 12 = 0 or x + 16 = 0 x = 12 or x = – 16 The side of square cannot be negative x – 16

Hence x = 12 x + 4 = 12 + 4 = 16

The side of smaller square is 12 m and bigger square is 16 m.

(xi) Exterior angle of a regular polygon having n-sides is more that of thepolygon having n2 sides by 500 . Find the number of the sides of eachpolygon. (5 marks)

Sol. Number of sides of one of the regular polygon = nNumber of sides of the other regular polygon = n2

Exterior angle of a polygon having ‘n’ sides = 360

n

o

Exterior angle of a polygon having ‘n2’ sides = 360

n

o

2


–n n2 = 50

360 1 1

–n n

2 = 50

n – 1

n2 =50

360

n – 1

n2 =5

36 36 (n – 1) = 5 (n2) 36n – 36 = 5n2

0 = 5n2 – 36n + 36 5n2 – 36n + 36 = 0 5n2 – 30n – 6n + 36 = 0 5n (n – 6) – 6 (n – 6) = 0 (n – 6) (5n – 6) = 0 n – 6 = 0 or 5n – 6 = 0 n = 6 or 5n = 6

n = 6 or n = 6

5


SCHOOL SECTION 153

‘n’ is the number of sides of any polygon

n 6

5Hence n = 6

n2 = (6)2 = 36

The number of sides of required regular polygons are 6 and36 cm respectively.

(xiv) The sides of one regular hexagon is larger than that of the other regularhexagon by 1cm . If the product of their areas is 243, than find thesides of both the regular hexangons. (5 marks)

Sol. Let the side of the smaller regular hexagon be x cm. The side of the bigger regular hexagon is (x + 1) cm

Area of regular hexagon = 3 3

2 × (side)2


3 3

2 x2 +

3 3

2 (x + 1)2 = 243

9 3

4

× x2 (x + 1)2 = 243

[x (x + 1)]2 = 243 × 4

9 3 [(x2 + x)]2 = 36

Taking square root on both the sides we get,x (x + 1) = + 6

x (x + 1) = – 6 is not acceptable because product of side of hexagon cannotbe negative

x (x + 1) = 6 x2 + x = 6 x2 + x – 6 = 0 x2 + 3x – 2x – 6 = 0 x (x + 3) – 2 (x + 3) = 0 (x + 3) (x – 2) = 0 x + 3 = 0 or x – 2 = 0 x = – 3 or x = 2 x is a side of regular hexagon x – 3

Hence x = 2 x + 1 = 2 + 1 = 3

The sides of both the regular hexagon are 2 cm and 3 cm respectively.

(xxii) Around a square pool there is a footpath of width 2km. if the area of the

footpath is 54

times that of the pool. Find the area of the pool.(5 marks)

Sol. Let the side of inner square i.e. pool be x m. The width of foot path around the pool is 2 m The side of outer square is (x + 2) m Area of square = side2


SCHOOL SECTION154

As per the given condition,(Area of outer square) = (Area of inner square) + (Area of footpath)

(x + 4)2 = x2 + 5

4 x2

x2 + 8x + 16 = x2 + 5

4x2

8x + 16 =5

4 x2

Multiplying throughout by 4 we get,32x + 64 = 5x2

5x2 – 32x – 64 = 0 5x2 – 40x + 8x – 64 = 0 5x (x – 8) + 8 (x – 8) = 0 (x – 8) (5x + 8) = 0 x – 8 = 0 or 5x + 8 = 0 x = 8 or 5x = – 8

x = 8 or x = – 8

5x =

– 8

5 is not acceptable because side of pool cannot be negative.

x = 8 x2 = 82

x2 = 64

Area of pool is 64 sq. m.

(ii) In garden there are some rows and columns. The number of trees in arow is greater than that in each column by 10. Find the number of treesin each row if the total number of trees are 200. (3 marks)

Sol. Let the number of trees in each column be x The number of trees in each column is x + 10

As per the given condition,x (x + 10) = 200

x2 + 10x – 200 = 0 x2 + 20x – 10x – 200 = 0 x (x + 20) – 10 (x + 20) = 0 (x + 20) (x – 10) = 0 x + 20 = 0 or x – 10 = 0 x = –20 or x = 10 The number of trees cannot be negative x – 20

Hencex = 10

x + 10 = 10 + 10 = 20

The number of trees in each row is 20.

WORD PROBLEMS BASED ON SPEED, DISTANCE AND TIME


30. Solve the following :(xx) From the same place at 7 am ‘A’ started walking in the north at the speed

of 5 km/hr. After 1 hour B started cycling in the east at a speed of16 km/hr. At what time they will be at distance of 52 km apart from eachother. (5 marks)

Sol. Let O be the point from where A and B started their journey.

2 km

2 km

(x+ 4)km

(x+ 4)km

2km

2km

footpath


SCHOOL SECTION 155

A reached at point N while B reached at point E and distance NE is52 kmLet the time taken by B to reach at point E be x hrs

The time taken by A to reach at point N (x + 1) hrs. Distance = Speed × Time

As per the given condition,In right angled NOE,

(NE)2 = (ON)2 + (OE)2 [By Pythagoras theorem] (52)2 = [5 (x + 1)]2 + (16x)2

2704 = (5x + 5)2 + (16x)2

0 = 25x2 + 50x + 25 + 256x2 – 2704 0 = 281x2 + 50x – 2679 = 0 281x2 – 843x + 893x – 2679 = 0 281x (x – 3) + 893 (x – 3) = 0 (x – 3) (281x + 893) = 0 x – 3 = 0 or 281x + 893 = 0 x = 3 or 281x = – 893

x = 3 or x = –893

281 x is the time taken by A

x –893

281Hence x = 3

x + 1 = 4 A took 4 hrs to reach at point N from point O

He started his journey at 7 a.m. He will be at point N at 11 a.m.

Hence B will be at point E at 11 a.m.

A and B will be at distance of 52 km apart from each other at 11 a.m.

(xvii) A car covers a distance of 240km with some speed is increased by 20km/hr, it will cover the same distance in 2 hours less. find the speed ofthe car. (5 marks)

Sol. Let the original speed of car be x km/hr. Distance covered is 240 km

Speed = Distance

Time

Time = Dis tance

Speed

Time taken by car =240

hrsx

New speed of car = (x + 20) km/hr

New time taken by car =240

hrsx 20


–x x 20 = 2

1 1

240 –x x 20

= 2

N

O E

52 km

5 (

x +

1)

km

16x km


SCHOOL SECTION156

x 20 – x

x (x 20)

=

2

240

20

x 20x2 =1

120 20 (120) = 1 (x2 + 20x) 2400 = x2 + 20x 0 = x2 + 20x – 2400 x2 + 20x – 2400 = 0 x2 + 60x – 40x – 2400 = 0 x (x + 60) – 40 (x + 60) = 0 (x + 60) (x – 40) = 0 x + 60 = 0 or x – 40 = 0 x = – 60 or x = 40 The speed of car can never be negative. x – 60

Hence x = 40

The original speed of car is 40 km/hr.

(v) A man riding on a bicycle cover a distance of 60 km in a direction ofwind and comes back to his original position in 8 hours. If the speed ofthe wind is 10 km/hr. Find the speed of the bicycle. (5 marks)

Sol. Let the speed of bicycle be x km / hrSpeed of wind is 10 km /hr

Speed of bicycle in the direction of wind = (x + 10) km/hrSpeed of the bicycle against the direction of wind = (x – 10) km/hr

Also, Speed = Dis tance

Time

Time = Dis tance

Speed

Time taken by man while riding in the direction of wind = 60

x 10

hrs

Time taken by man while riding against the direction of wind = 60

x – 10

hrs


60 60

x 10 x – 10

= 8

1 1

60x 10 x – 10

= 8

x – 10 x 10

(x 10) (x – 10)

=

8

60

2x

x – 1002 =2

15Dividing throughout by 2 we get,

x

x – 1002 =1

15 x2 – 100 = 15x x2 – 15x – 100 = 0 x2 – 20x + 5x – 100 = 0 x (x – 20) + 5 (x – 20) = 0 (x – 20) (x + 5) = 0


SCHOOL SECTION 157

x – 20 = 0 or x + 5 = 0 x = 20 or x = – 5 The speed of bicycle cannot be negative x – 5

Hence x = 20

The speed of bicycle is 20 km / hr.

(xviii) An express train takes 30 min less for a journey of 440 km, if its usualspeed is increased by 8 km/hr. find its usual speed. (5 marks)

Sol. Let the usual speed of train be x km/hr. Distance = Time × Speed

Speed = Dis tance

Time

Time = Dis tance

SpeedDistance covered by train = 440 km

Time taken by train =440

x

hrs

New speed of train = (x + 8) km/hr

New time taken by train = 440

x 8

hrs.


–x x 8 =

1

2

130 min hr

2

1 1

440 –x x 8

=1

2

x 8 – x

x (x 8)

=

1 1

2 440

8

x 8x2 =1

880 8 (880) = 1 (x2 + 8x) 7040 = x2 + 8x 0 = x2 + 8x – 7040 = 0 x2 + 88x – 80x – 7040 = 0 x (x + 88) – 80 (x + 88) = 0 (x + 88) (x – 80) = 0 x + 88 = 0 or x – 80 = 0 x = – 88 or x = 80 The speed of train can never be negative. x – 88

Hence x = 80

The usual speed of train is 80 km/hr.

WORD PROBLEMS BASED ON COST

EXERCISE - 2.10 (TEXT BOOK PAGE NO. 63) :13. The cost of bananas is increased by Re. 1 per dozen, one can get 2 dozen less

for Rs. 840. Find the original cost of one dozen of banana. (5 marks)Sol. Let the cost of banana per dozen be Rs. x.

Amount for which bananas are bought = Rs. 840


SCHOOL SECTION158

No. of dozens of bananas for Rs 840 is 840

xNew cost of banana per dozen = Rs. (x + 1)

New No. of dozens of bananas for Rs 840 = 840

x 1As per the given condition,

840 840–

x x 1 = 2

1 1

840 –x x 1

= 0

x 1 – x

840x (x 1)

= 2

1

840x x

2 = 2

840 = 2 (x2 + x) 2x2 + 2x – 840 = 0

Dividing throughout by 2, 2x2 + x – 420 = 0 x2 – 20x + 21x – 420 = 0 x (x – 20) + 21 (x – 20) = 0 x – 20 = 0 or x + 21 = 0 x = 20 or x = –21 The cost of bananas cannot be negative. x –21

Hence x = 20

The original cost of one dozen banana is Rs. 20.

WROD PROBLEMS BASED ON WORK

PROBLEM SET - 2 (TEXT BOOK PAGE NO. 167) :30. Solve the following :(vii) For doing some work Ganesh takes 10 days more than John. If both

work together they complete the work in 12 days. Find the number ofdays if Ganesh worked alone? (5 marks)

Sol. Let the number of days required by John alone to complete thework be x days and Ganesh alone is (x + 10) days

No. of days requred by ganesh along is (x + 10 ) days.Also number of days required by both to complete the same work is12 days

Work done by John in 1 day = 1

x

Work done by Ganesh in 1 day = 1

x + 10

Work done by both in 1 day = 1

12As per the given condition,

1 1+

x x + 10 =1

12

x + 10 + x

x (x + 10) =1

12


SCHOOL SECTION 159

2x + 10

x + 10x2 =1

12 12 (2x + 10) = 1 (x2 + 10x) 24x + 120 = x2 + 10x 0 = x2 + 10x – 24x – 120 x2 – 14x – 120 = 0 x2 – 20x + 6x – 120 = 0 x (x – 20) + 6 (x – 20) = 0 (x – 20) (x + 6) = 0 x – 20 = 0 or x + 6 = 0 x = 20 or x = – 6 The numbers of days cannot be negative x – 6

Hence x = 20 x + 10 = 20 + 10 = 30

Ganesh alone worked for 30 days.

(xii) Tinu takes 9 days more than his father to do a certain piece of work.Together they can do the work in 6 days. How many days will tinu taketo do that work. (5 marks)

Sol. Let the number of days required by father alone to do a certainpiece of work be ‘x’ days and by Tinu alone is (x + 9) days

No. of days required by tinu along is (x + 9) daysAlso number of days required by both to complete the same work is 6 days.

Work done by father in 1 day = 1

x

Work done by Tinu in 1 day = 1

x 9+

Work done by both in 1 day = 1


1 1

x x 9+

+ =1

6

x 9 x

x (x 9)

+ +

+ =1

6

2x 9

x 9x

+

+2 =1

6 6 (2x + 9) = 1 (x2 + 9x) 12x + 54 = x2 + 9x 0 = x2 + 9x – 12x – 54 0 = x2 – 3x – 54 x2 – 3x – 54 = 0 x2 – 9x + 6x – 54 = 0 x (x – 9) + 6 (x – 9) = 0 (x – 9) (x + 6) = 0 x – 9 = 0 or x + 6 = 0 x = 9 or x = – 6 The number of days cannot be negative x 6

Hence x = 9 x + 9 = 9 + 9 = 18

Tinu alone requires 18 days to complete the work.


SCHOOL SECTION160

(xxi) One tank can be filled up by two taps in 6 hours. The smaller tap alonetakes 5 hours more than the bigger tap alone. Find the time required byeach tap to fill the tank separately. (5 marks)

Sol. Let the time taken to fill a tank by a bigger tap alone be x hrs. The time taken by smaller tap alone is (x + 5) hrs.

Time taken by both the taps together to fill the same tank is 6 hrs.

Portion of tank filled in 1 hr by bigger tap = 1

x

Portion of tank filled in 1 hr by smaller tap = 1

x 5

Portion of tank filled in 1 hr by both taps together = 1


1

x +

1

x 5 =1

6

x 5 x

x (x 5)

=

1

6

2x 5

x 5x

2 =

1

6 6 (2x + 5) = 1 (x2 + 5x) 12x + 30 = x2 + 5x 0 = x2 + 5x – 12x – 30 x2 + 3x – 10x – 30 = 0 x (x + 3) – 10 (x + 3) = 0 (x + 3) (x – 10) = 0 x + 3 = 0 or x – 10 = 0 x = – 3 or x = 10 x is the time taken bybigger tap x – 3

Hence x = 10 x + 5 = 10 + 5 = 15

Time taken by bigger tap alone is 10 hrs and smaller tap alone is 15 hrs.

MCQ’s1. The sum of roots ( + ) = ............. .

(a)b

–a

(b)b

a

(c)c

–a

(d)c

a

2. In formula method the value of x = ............. .

(a)b – b – 4ac

2a

2

(b)– b b – 4ac

2a

2

(c)b b – 4ac

2a

2

(d)b 4ac

2a

2

3. In complemiting square method the formula of third term is ............. .

(a) 2 × (coefficient of x2) (b)1

2 × (coefficient of x2)

(c)1

coefficient of x2

2

(d) (2 × coefficient of x)2


SCHOOL SECTION 161

4. The product of the roots ( . ) = ............. .

(a)b

–a

(b)c

–a

(c)b

a(d)

c

a

5. The solution set of y2 – 16y + 63 = 0 are = ............. .(a) – 9 and – 7 (b) – 9 and 7(c) 9 and – 7 (d) 9 and 7

6. The factors of y2 – 5y – 24 = 0 is = ............. .(a) – 8 and – 3 (b) – 8 and 3(c) 8 and – 3 (d) 8 and 3

7. The roots of the quadratic equation x2 + 5x – 14 is = ............. .(a) 2 (b) – 7(c) – 3 (d) – 2

8. Which of these following is quadratic equation ............. .

(a)5

m – 4m 5m

(b) n3 – n + 4 = n3

(c)5

– 3 xx

2(d) 13 = – 5y2 – y3

9. If the roots of quadratic equation are real and equal then must be ............. .(a) equal to zero (b) greater then zero(c) greater than one (d) equal to one

10. If m2 – 36 = 0 then m2 is equal to ............. .(a) 6 (b) 36(c) 9 (d) – 6

11. Three times the square of natural number is 363 is written in themathematical equation form as ............. .(a) x2 + 3 = 363 (b) x2 – 3 = 363

(c) 3x2 = 363 (d)x

3633

2

12. Which of these following is not an quadratic equation ............. .

(a)5

– x 2x 93

2(b) (x + 3) (x + 4) = 0

(c)5

– 3 xx

2(d) m –

5

m = 4m + 5

13. If one root of quadratic equation is kx2 – 7x + 12 = 0 is 3 then value of k is............. .(a) – 1 (b) 1(c) 3 (d) none of these

14. If one root of quadratic equation is 1 – 3 then product of root is ............. .

(a) 2 (b) – 2(c) – 44 (d) 44


SCHOOL SECTION162

15. The quadratic equation is in form of ............. .(a) x2 + (sum of roots)x – (product of roots) = 0(b) x2 – (sum of roots)x + (product of roots) = 0(c) x2 + (product of roots)x – (sum of roots) = 0(d) x2 – (product of roots) x + (sum of roots) = 0

16. If one root of quadratic equation is 4 2 – 3then sum of roots is ............. .

(a) –8 2 (b) 8 2(c) 23 (d) – 6

17. If sum of natural number and its square is 30, then quadratic equation iswritten as ............. .(a) x2 – x – 30 = 0 (b) x2 + x – 30 = 0(c) x2 – x + 30 = 0 (d) x2 + x + 30 = 0

18. x2 + 1

x2 = ............. .

(a)1

x – – 2x

2

2 (b)1

x 2x

2

2

(c)1

x – 2x

2

(d) none of these

19. 3 + 3 = ............. .(a) ( + )3 – 3 ( + ) (b) ( – )3 + 3 ( – )(c) ( + )3 – 3 ( – ) (d) ( – )3 – 3 ( – )

20. ( – )2 = ............. .(a) ( + )2 + 4 (b) ( + )2 – 3(c) ( + )2 – 4 (d) ( + )2 + 2

: ANSWERS :

1. (a)b

–a

2. (b) – b b – 4ac

2a

2

3. (c)1

coefficient of x2

2

4. (d) c

a5. (d) 9 and 7 6. (b) – 8 and 3

7. (b) – 7 8. (a) 5

m – 4m 5m

9. (a) equal to zero 10. (b) 36

11. (c) 3x2 = 363 12. (c) 5

– 3 xx

2

13. (b) 1 14. (b) – 2

15. (b) x2 – (sum of roots)x + (product of roots) = 0

16. (d) – 6 17. (b) x2 + x – 30 = 0

18. (c)1

x – 2x

2

19. (a) ( + )3 – 3 ( + )

20. (c) ( + )2 – 4


SCHOOL SECTION 163

HOTS PROBLEMS(Problems for developing Higher Order Thinking Skill)

10. If the roots of the quadratic equation ax2 + cx + c = 0 are in the ratio

p : q show that p q c

+ + = 0q p a (3 marks)

Sol. As we know, as per the given equation

+ =–c

a

=c

a

Also, =

p

q

Now,p q c

q p a = .

= .

=+

.

=+ +

=

–c ca a

=0

= 0

p q c

0q p a

Hence proved.

11. If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal tothe sum of the squares of their reciprocals then prove that 2a2c = c2b +b2a. (4 marks)

Ans. Let the root of the quadratic equation ax2 + bx + c = 0 be and and as we know for the ab are equation

+ =– b

a

=c

aAs per the given condition,

+ =1 1

2 2

+ =2 2

2 2

+ = 2 – 2 2 2 2

2


SCHOOL SECTION164

+ =

– 2 2

2

– b

a=

b c– – 2

a a

ca

2

2

b c

–a a

2

2 =b 2c

–aa

2

2

bc

–a

2

3 =b 2ac

–a a

2

2 2

bc

–a

2

3 =b – 2ac

a

2

2

Multiplying both sides by a3, we get

bca –

a

23

3 =b – 2ac

a –a

23

2

– bc2 = a (b2 – 2ac)– bc2 = ab2 – 2a2c

2a2c = c2b + b2aHence proved.

12. Form the quadratic equation whose roots are the squares of the sum and squareof the difference of the equation 2x2 + 2(m + n)x + m2 + n2 = 0 (4 marks)

Ans. Let and be the roots of the equation2x2 + 2 (m + n)x + m2 + n2 = 0 as we know

+ =–2 (m n)

2

and =

m n

2

2 2

Let roots of new quadratic equation be 1 and 1

1 = ( + )2 = (m + n)2 and . = m n

2

2 2

( – )2 = ( + )2 – 4

= m2 + 2mn + n2 – 4 m n

2

2 2

= m2 + 2mn + n2 – 2m2 – 2n2

= – m2 + 2mn – n2

= – (m2 – 2mn + n2)1 = (a – b)2 = – (m – n)2

Now, 1 + 1 = m2 + 2mn + n2 – m2 + 2mn – n2

1 + 1= 4mn1 . 1 = (m + n)2 × – (m – n)2 = – (m2 – n2)2

Quadratic equation is x2 – 4mnx – (m2 – n2)2 = 0.

13. Find the condition that the equations ax2 + bx + c = 0 and a1x2 + b1x + c1 = 0

may have a common root. Find this common root. (5 marks)Sol. ax2 + bx + c = 0 ......(i)

a1x2 + b1x + c1 = 0 .....(ii)

Multiplying (i) by a1,a1ax2 + a1bx + a1c = 0 ......(iii)Multiplying (ii) by a,a1ax2 + ab1x + ac1 = 0 ......(iv)


SCHOOL SECTION 165

Subtracting (iv) from (iii),a1ax2 + a1bx + a1c = 0a1ax2 + ab1x + ac1 = 0

(–) (–) (–)a1bx – ab1x + a1c – ac1 = 0

x (a1b – ab1) = ac1 – a1c

x = ac – a c

a b – ab1 1

1 1

Substituting x = ac – a c

a b – ab1 1

1 1 in (i),

ac – a c ac – a ca b c

a b – ab a b – ab

2

1 1 1 1

1 1 1 1 = 0

a ac – a c b ac – a c

ca b – aba b – ab

2

1 1 1 12

1 11 1

= 0

Multiplying throughout by (a1b – ab1)2 we get,

a (ac1 – a1c)2 + b (ac1 – a1c) (a1b – ab1)2 + c(a1b – ab1)

2 = 0The condition for the equations to have common root is

a (ac1 – a1c)2 + b (ac1 – a1c) (a1b – ab1)2 + c (a1b – ab1)

2 = 0

14. Find the value if p of the equations 3x2 – 2x + p and 6x2 – 17x + 12 = 0have a common root. (4 marks)

Sol. 6x2 – 17x + 12 = 0 6x2 – 9x – 8x + 12 = 0 3x (2x – 3) – 4 (2x – 3) = 0 (2x – 3) (3x – 4) = 0 2x – 3 = 0 or 3x – 4 = 0

x = 3

2or x =

4

3The first equation 3x2 – 2x + p = 0If x = 3 is a common root then this value of ‘x’ will satisfy the equation

3 3 3

– 2 p2 2

2

= 0

9

3 – 3 p4

= 0

9

– 3 p2

= 0

3

p2 = 0

p =3

–2

15. If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equalto the sum of the squares of their reciprocals. Show that bc2, ca2, ab2

are in A.P. (5 marks)Sol. To show bc2, ca2, ab2 are in A.P.

i.e. to prove ca2 – bc2 = ab2 – ca2

i.e. ca2 + ca2 = ab2 + bc2

i.e. 2ca2 = ab2 + bc2 .......(i)Let the roots quadratic equation ax2 + bx + x = 0 be a and b as weknow for the above equation


SCHOOL SECTION166

+ =– b

a

=c

aAs per the given condition,

+ =1 1

2 2

+ =2 2

2 2

+ = 2 – 2 2 2 2

2

+ =

– 2 2

2

– b

a=

– b c– 2

a a

ca

2

2

– b c

a a

2

2 =b 2c

–ac

2

2

– bc

a

2

3 =b 2ac

–a a

2

2 2

– bc

a

2

3 =b – 2ac

a

2

2

Multiplying both sides by a3 we get,

– bca

a

23

3 =b – 2ac

aa

23

2

– bc2 = a (b2 – 2ac) – bc2 = ab2 – 2a2c – bc2 = ab2 – a2c – a2c a2c – bc2 = ab2 – a2c

ca2 – bc2 = ab2 – ca2

bc2, ca2, ab2 are in A.P.

34. Solve : 216x6 – 793x3 + 216 = 0 (5 marks)Sol. 216x6 – 793x3 + 216 = 0

216 (x3)2 – 793x3 + 216 = 0Substituting x3 = a we get,

216a2 – 793a + 216 = 0 216a2 – 64a – 729a + 216 = 0 216a2 – 64a – 729a + 216 = 0 8a (27a – 8) – 27 (27a – 8) = 0 (27a – 8) (8a – 27) = 0 27a – 8 = 0 or 8a – 27 = 0 27a = 8 or 8a = 27

a = 8

27or a =

27

8


SCHOOL SECTION 167

Resubstituting a = x3

x3 =8

27or x3 =

27

8Taking cube roots throughout,

x =2

3or x =

3

2

Solution set = 2 3

,3 2

35. A man travels by boat 36 km down a river and back in 8 hours. If thespeed of his boat in still water is 12 km per hour, find the speed of theriver curved. (4 marks)

Sol. Speed of boat in still water = 12 km/hr.Let speed of river current = x km/hr.

Speed of boat up the river = (12 – x) km/hr.Speed of boat down the river = (12 + x) km/hr.

Time = Distance

Speed

Time taken by boat to travel 36 km down the river = 36

12 x

hrs.


36 36

12 x 12 – x+

+ = 8 1 1

3612 x 12 – x

= 8

12 – x 12 x

36(12 x) (12 – x)

= 8

24

36144 – x

2 = 8

36 × 24

8= 144 – x2

108 = 144 – x2

x2 = 144 – 108 x2 = 36 x = + 6 [Taking square roots]

x = – 6 is not acceptable because speed cannot be negative.

38. A business man bought some items for Rs. 600, keeping 10 items forhimself he sold the remaining items at a profit of Rs. 5 per item. Fromthe amount received in this deal he could buy 15 more items. Find theoriginal price of each item. (5 marks)

Sol. Let the original price of each item be Rs. xTotal cost price = Rs. 600

No. of items bought = 600

x

No. of items sold after keeping 10 items = 600

– 10x

Selling price each item = Rs. (x + 5)

Total selling price = (x + 5) 600

– 10x


SCHOOL SECTION168

Net profit made (x + 5) 600

– 10x

– 600

But profit is equal to cost of 15 items = Rs. xAs per the given condition,

(x + 5) 600

– 10x

– 600 = 15x

600

x – 10x

+ 600

– 10x

– 600 = 15x

600 – 10x + 3000

x – 50 – 600 = 15x

– 10x + 3000

x – 50 – 15x = 0

– 25x + 3000

x – 50 = 0

Multiplying throughout by x,– 25x + 3000 – 50x = 0

25x2 – 50x + 3000 = 0Dividing throughout by – 25 we get,

x2 + 2x – 120 = 0 x2 + 12x – 10x – 120 = 0 x (x + 12) – 10 (x + 12) = 0 (x + 12) (x – 10) = 0 x + 12 = 0 or x – 10 = 0 x = – 12 or x = 10

x = – 12 is not acceptable because cost cannot be negative. x = 10

Original price of each items is Rs. 10.

CHAPTER 2 : Quadratic Equations

SET - A

S.S.C.

ALGEBRA

Marks : 30

Duration : 1 hr.

Q.1. Atttempt any TWO of the following : 2

(i) State whether the following equation is a (y – 2) (y + 2) = 0.

(ii) Write the following quadratic equations in standard form : (m + 4)(m – 10) = 0.

(iii) Find the value of discriminant of of the following equation :3x2 + 2x – 1 = 0.

Q.2. Attempt any TWO of the following : 4

(i) If one root of the quadratic equation x2 – 7x + k = 0 is 4, then findthe value of k.

(ii) Solve the following quadratic equation by factorization method :x2 – 13x – 30 = 0.

(iii) Solve the following quardratic equation by factorization method :y2 – 3 = 0.


(i) Solve the following quadratic equation by completing square :x2 + 8x + 9 = 0.

(ii) Form the quadratic equation if its roots are – 2 and 11

2.

(iii) Find the value of k for which given equation has real and equal roots :(k – 12)x2 + 2 (k – 12)x + 2 = 0


(i) Find m, if the roots of the quadratic equation (m – 1) x2 – 2(m – 1) x + 1 = 0 has real and equal roots.

(ii) If one root of the quadratic equation kx2 – 5x + 2 = 0 is 4 times theother, find k.

(iii) Solve the following equation : x2 + 12

x2 = 7.


(i) Solve the following equation : 1 1

2 x + – 9 x + + 14 = 0x x

22

(ii) The product of four consecutive positive integers is 840. find thelargest number.

(iii) If the difference of the roots of the quadratic equation is 3 and differencebetween their cubes is 189, find the quadratic equation.

Best Of Luck

CHAPTER 2 : Quadratic Equations

SET - A

S.S.C.

ALGEBRA

Marks : 30

Duration : 1 hr.

Q.1. Atttempt any TWO of the following : 2

(i) State whether following is a quadratic equation 3y2 – 7 = 3 y.

(ii) Write the following quadratic equation in standard form (m + 4)(m – 10) = 0.

(iii) Find the value of discriminant of each of the following equation :3x2 + 2x – 1 = 0.


(i) Solve the following quadratic equation by factorization method :x2 + 10x + 24 = 0.

(ii) If one root of the quadratic equation 3y2 – ky + 8 = 0 is 2

3, then find

the value of k.

(iii) Solve the following quadratic equation by factorization method :49x2 = 36.


(i) Solve the following quadratic equation using formula :x2 + 3x – 10 = 0.

(ii) Solve the following quadratic equation using formula :2x2 + 5x – 2 = 0.

(iii) From the quadratic equation whose roots are 3

4 and

– 2

3.


(i) Find the value of k for which given equation has real and equalroots : (k – 12)x2 + 2 (k – 12)x + 2 = 0

(ii) Find k, if the roots of the quadratic equation x2 + kx + 40 = 0 are inthe ratio 2 : 5.

(iii) Solve : 35y2 + 12

y2 = 44.


(i) Solve the following equation :

1 112 x + – 56 x + + 89 = 0

xx

22

(ii) A rectangular playground is 420 sq.m. If its length is increases by7 m and breadth is decreased by 5 metres, the area remains thesame. Find the length and breadth of the playground ?

(iii) If the difference of the roots of the quadratic equation is 5 and thedifference of their cubes is 215, find the quadratic equation.

Best Of Luck

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