123

After learning this chapter you will be able to :* Identify a quadratic equation.* Distinguish between a pure and an adfected quadratic equation.* Solve simple problems on pure and adfected quadratic equations.* Identify the standard form of a quadratic equation.* Solve the quadratic equations by factorization.* Solve the quadratic equations by using formula.* Find the value of the discriminant and know the nature of the roots.* Frame the quadratic equation for the given roots.* Solve the quadratic equation graphically.

QUADRATIC EQUATIONSYou are familiar with the properties and area of a square and a rectangle.Consider a square of side x units and its perimeter 16 units, then the length

of the side isPerimeter = 4 (length of the side)

16 = 4xor

4x = 16 .............. (1)

In equation (1) the degree of the variable is one.

Can you recall the name of such an equation?Such an equation is a linear equation.

4x = 16

x = 416

x = 4

An equation involving avariable of degree one onlyis a linear equation.

A Linear equation hasonly one root.

x

x

xx

5 QUADRATIC EQUATIONS

124

An equation involving avariable of degree 2 isquadratic equation.

a

a

aa

m m

(m+2)

(m+2)

Consider a square of side a units and its area 25 square units.Area of the square = (side)2

25 = a2

or

a2 = 25 .............. (2)

In equation (2) the degree of the variable is two

What do you call such an equation?Such an equation is a quadratic equation.

a2 = 25 a = 5

a = + 5 or a = 5

Consider a rectangle of sides m and (m + 2) units and its area is 8 sq units. Area of a rectangle = (length) (breadth)

8 = (m) (m + 2) 8 = m2 + 2m

or

m2 + 2m = 8 .......................... (3)Compare the equation (2) and (3)In equation (2) a2 = 25, variable occurs only in second degree.In equation (3) m2 + 2m = 8, variable occurs in second degree as well as in first degree.

Quadratic equation involving a variable only in second degree is aPure Quadratic Equation.

Example :(1) x2 = 9 (2) 2a2 = 18

If the terms in the RHS are transposed to LHS then,(1) x2 9 = 0 (2) 2a2 18 = 0

An equation that can be expressed in the form ax2 + c = 0, where a andc are real numbers and a 0 is a pure quadratic equation.

A Quadratic equation hasonly two roots.

125

Quadratic equation involving a variable in second degree as well as in firstdegree is an Adfected Quadratic Equation

Example :(1) x2 + 3x = 10 (2) 3a2 a = 2

If the terms in the RHS are transposed to LHS then,(1) x2 + 3x 10 = 0 (2) 3a2 a 2 = 0

ax2 + bx + c = 0 is the standard form of a quadratic equation where a, band c and variables and a 0.

1. Solving Pure Quadratic equationExample 1 : Solve the equation 3x2 27 = 0

Solution : 3x2 27 = 0 3x2 = 27

x = 327

x2 = 9

x = 9x = +3 or x = 3

Example 2 : Solve the equation 4y2 9 = 0Solution : 4y2 9 = 0

4y2 = 9

y2 = 49

y = 49

y = 23

y = 23

+ or y = 23

126

Example 3 : Solve the equation 99 = 4r2 1Solution : 99 = 4r2 1

4r2 1 = 99

4r2 = 99 + 1 4r2 = 100

r2 = 4100

= 25

r = 25

r = 5r = +5 or r = 5

Example 4 : Solve the equation (m + 8)2 5 = 31Solution : (m + 8)2 5 = 31

(m + 8)2 = 31 + 5(m + 8)2 = 36

(m + 8)2 = 36

(m + 8) = 36 m = 8 6m = 8 + 6 or m = 8 6

m = 2 or m = 14

Example 5 : If A = 2r ; Solve for r.

Solution : A = 2r

2r = A

r2 =

A

r =

A

127

l2

Example 6 : If l2 = r2 + h2. Solve for h and find the value of h if l = 15 and r = 9.Solution : l2 = r2 + h2

or

r2 + h2 = l2

h2 = l2 r2

h = 22 rl

h = 22 915 (substituting l = 15, r = 9)h = 81225

h = 144 h = 12 h = +12 or h = 12

Example 7 : If B = 4a.3 2

Solve for a and find the value of a if B = 16 3 .

Solution : B = 4a.3 2

a2 = 3B4

a = 3B4 (Substituting B = 16 3 )

a = 33164

//

a = 64 , a = 8

a = + 8 or a = 8

Exercise : 5.1

A. Classify the following equations into pure and adfected quadratic equation.1) x2 + 2 = 6 2) a2 + 3 = 2a 3) p (p 3) = 1

4) 2m2 = 72 5) k2 k = 0 6) 7y = y35

128

If mn = 0, then eitherm = 0 or n = 0

B. Solve the equations

1) 5x2 = 125 2) m2 1 = 143 3) 4a = a

81

4) 2

2x

43

= 417 5) (2m 5)2 = 81 6)

18)4( 2x

= 92

C.

1) If A = 2 2r Solve for r and find the value of r if A = 77 and = 722

2) If V = hr2 Solve for r and find the value of r if V = 176 and h = 143) If r2 = l2 + d2 Solve for d and find the value of d if r = 5 and l = 4.4) If c2 = a2 + b2 Solve for b. If a = 8 and c = 17 and find the value of b.5) If K = 1/2mv2 Solve for v and find the value of v if K = 100 and m = 26) If v2 = u2 + 2as. Solve for v. If u = 0, a = 2 and s = 100, find the value

of v.

2. Solving the adfected quadratic equation by factorization :Example 1 : Solve the quadratic equation a2 3a + 2 = 0

Solution : a2 3a + 2 = 0i. Resolve the expression a2 2a 1a + 2 = 0ii. Factorize a(a 2) 1 (a 2) = 0iii. Taking the common factor (a 2) (a 1) = 0iv. Equate each factor to zero a 2 = 0 or a 1 = 0v. The roots are a = 2 or a = 1

Example 2 : Solve the quadratic equation m2 m = 6Solution : m2 m = 6

m2 m 6 = 0m2 3m + 2m 6 = 0m(m 3) +2 (m 3) = 0(m 3) (m + 2) = 0

Either (m 3) = 0 or (m + 2) = 0m = +3 or m = 2

129

x2

1x 1x

Example 3 : Solve the quadratic equation 2x2 3x + 1 = 0Solution : 2x2 3x + 1 = 0

2x2 2x 1x + 1 = 02x (x 1) 1 (x 1) = 0

(x 1) (2x 1) = 0Either (x 1) = 0 or (2x 1) = 0

x = 1 or x = 21

Example 4 : Solve the quadratic equation 4k (3k 1) = 5.Solution : 4k (3k 1) = 5

12k2 4k 5 = 012k2 10k + 6k 5 = 02k (6k 5) + 1(6k 5) = 0

(6k 5) (2k + 1) = 0Either (6k 5) = 0 or (2k + 1) = 0

k = 65

or k = 21

Exercise : 5.2A. Find the roots of the following equations

1) x(x 3) = 0 2) a (a + 5) = 0 3) m2 4m = 04) 3k2 + 6k = 0 5) (y + 6) (y + 9) = 0 6) (b 3) (b 5) = 07) (2n + 1) (3n 2) = 0 8) (5z 2) (7z + 3) = 0

B. Solve the quadratic equations1) x2 + 15x + 50 = 0 2) a2 5a + 6 = 0 3) y2 = y + 24) 6 p2 = p 5) 30 = b2 b 6) 2x2 + 5x 12 = 07) 6y2 + y 15 = 0 8) 6a2 + a = 5 9) 13m = 6(m2 + 1)10) 0.2t2 0.04t = 0.03

Consider the equation x2 + 3x + 1 = 0It cannot be factorised by splitting the middle term.How do you solve such an equation ?

It can be solved by using Formula.

130

3. Solving quadratic equation by formula methodGeneral form of a quadratic equation ax2 + bx + c = 0

Divide by a 0a

c

a

bxa

ax2=++

Transpose the constant term to R.H.S.a

c

a

bxx

2=+

Add 2

a2b

to both the sides

222

a2b

a

c

a2b

a

bxx

+=

++

2

22

a4b

a

c

a2b

x +=

+

2

22

a4bac4

a2b

x+

=

+

Simplify 222

a4ac4b

a2b

x

=

+

Taking square root 22

a4ac4b

a2b

x=+

a2ac4b

a2b

x2

=+

a2ac4b

a2b

x2=

Roots area2

ac4bbx

2

=

a2

ac4bbx

2+

= or

a2ac4bb

x2

=

Roots of the equation ax2 + bx + c = 0 are x = a2

ac4bb 2

Note : The roots of the equation ax2 + bx + c = 0 can also be found usingSridharas method.

x x

xx

xx

x

x

x

x

x x

x

x

x

131

Example 1 : Solve the equation x2 7x + 12 = 0consider x2 7x + 12 = 0

This is in the form ax2 + bx + c = 0

the coefficients are a = 1, b = 7 & c = 12

The roots are given by x = a2

ac4bb 2

Substituting the values a = 1, b = 7 and c = 12

1x2)12)(1(4)7()7(

x

2

=

248497

x

=

Simplify x = 2

17

x = 217 +

or x = 217

x = 28

or x = 26

Roots are x = 4 or x = 3

Example 2 : Solve the equation 2p2 p = 15Consider 2p2 p = 15

2p2 p 15 = 0This is in the form ax2 + bx + c = 0The coefficients are a = 2, b = 1 and c = 15

The roots are given bya2

ac4bbx

2

=

Substituting the values a = 2, b = 1 and c = 15

p = )2(2)15)(2(4)1()1( 2

x

x

x

x

x x

x

x x

132

x

p = 4

12011 ++

p = 41211

p = 4111

p = 4111+

or p = 4111

p = 412

or p = 410

p = 3 or p = 25

Example 3 : Solve the equation 2k2 2k 5 = 0Consider 2k2 2k 5 = 0

This is in the form ax2 + bx + c = 0The coefficients are a = 2, b = 2 and c = 5

The roots are given by x = a2

ac4bb 2

Substituting the values a = 2, b = 2 and c = 5

k = )2(2)5)(2(4)2()2( 2

k = 4

4042 + =

4442

k = 4

1122 =

( )4

1112

The roots are k = 2111+

or k = 2

111

133

Example 4 : Solve the equation m2 2m = 2Consider m2 2m = 2

m2 2m 2 = 0This is in the form ax2 + bx + c = 0

Comparing the coefficients a = 1, b = 2 and c = 2

The roots are given by x = a2

ac4bb 2

m = )1(2)2)(1(4)2()2( 2

m = 2

842 ++

m = 2

122

m = ( )

2312

m = 31+ or m = 31Exercise : 5.3Solve the following equations by using formula

1) a2 2a 4 = 0 2) x2 8x + 1 = 0 3) m2 2m + 2 = 04) k2 6k = 1 5) 2y2 + 6y = 3 6) 8r2 = r + 2 7) p = 5 2p28) 2z2 + 7z + 4 = 0 9) 3b2 + 2b = 2 10) a2 = 4a + 6

4. Equations reducible to the form ax2 + bx + c = 0Example 1 : Solve the equation (x + 6) (x + 2) = x

Solution :(x + 6) (x + 2) = xx2 + 6x + 2x + 12 = xx2 + 8x + 12 x = 0x2 + 7x + 12 = 0x2 + 4x + 3x + 12 = 0x(x + 4) + 3 (x + 4) = 0(x + 4) (x + 3) = 0

Either (x + 4) = 0 or (x + 3) = 0x = 4 or x = 3

x

134

Example 2 : Solve the equation (a 3)2 + (a + 1)2 = 16Solution : (a 3)2 + (a + 1)2 = 16

Using (a + b)2 = a2 + 2ab + b2(a b)2 = a 2ab + b2

[(a)2 2(a)(3) + 32] + [a2 + 2 (a) (1) + 12] = 16a2 6a + 9 + a2 + 2a + 1 16 = 02a2 4a 6 = 0a2 2a 3 = 0a2 3a + 1a 3 = 0a(a 3) + 1 (a 3) = 0(a 3) (a + 1) = 0

Either (a 3) = 0 or (a + 1) = 0 a = 3 or a = 1

Example 3 : Solve 5(p 2)2 + 6 = 13 (p 2)Solution : 5(p 2)2 + 6 = 13 (p 2)Let p 2 = bthen 5b2 + 6 = 13b

5b2 13b + 6 = 05b2 10b 3b + 6 = 05b (b 2) 3 (b 2) = 0(b 2) (5b 3) = 0

Either (b 2) = 0 or (5b 3) = 0

b = 2 or b = 53

p 2 = 2 or p 2 = 53

( b = p 2)

p = 2 + 2 or p = 53

+ 12

p = 4 or p = 513

135

Example 4 : Solve the equation 1k1k

5k22k3

+=

+

+

Consider 1k1k

5k22k3

+=

+

+

Cross multiplying (3k + 2) (k 1) = (2k + 5) (k + 1)3k2 + 2k 3k 2 = 2k2 + 5k + 2k + 53k2 1k 2 2k2 7k 5 = 03k2 1k 2 2k2 7k 5 = 0

On simplification k2 8k 7 = 0This is in form of ax2 + bx + c = 0

The co-efficients are a = 1, b = 8, c = 7

The roots of the equationa2

ac4)b(bx

2+

=

k =

1x2)7)(1(4)8()8( 2

k =

228648 ++

k = 2

928

k = 2

2328 =

( )2

2342

k = 234

Example 5 : Solve the equation 1y23

4y

=

Consider 1y23

4y

=

Taking L.C.M. 1y46y2=

By cross multiplication y2 6 = 4yy2 4y 6 = 0

x

136

this is in the form ax2 + bx + c = 0comparing coefficients a = 1, b = 4, c = 6

y = )1(2)6)(1(4)4()4( 2

the roots of the equation are = 2

24164 +

y = 2

404

y = 2

1024 =

( )2

1022

y = 102+ or y = 102

Example 6 : Solve 1m24

3m1

2m4

+=

+

+

1m24

)3m)(2m()2m(1)3m(4

+=

++

++

1m24

6m3m2m2m12m4

2 +=

+++

+

1m24

6m5m10m3

2 +=

++

+

On Cross multiplying, 4m2 + 20m + 24 = 6m2 + 20m + 3m + 104m2 + 20m + 24 6m2 23m 10 = 02m2 3m + 14 = 0

This is in the Standard form 2m2 + 3m 14 = 02m2 + 7m 4m 14 = 0m(2m + 7) 2 (2m + 7) = 0(2m + 7) (m 2) = 0

Either (2m + 7) = 0 or (m 2) = 0

m = 27

or m = 2

137

Exercise : 5.4

A. Solve the following equations1) (x + 4) (x 4) = 6x 2) 2(a2 1) = a (1 a)3) 3(b 5) (b 7) = 4 (b + 3) 4) 8(s 1) (s + 1) + 2 (s + 3) = 15) (n 3)2 + n (n + 1)2 = 16 6) 11(m + 1) (m + 2) = 38 (m + 1) + 9m

B. Solve

1) 525

283

+

=

x

x

x

x 2) 1a7

1a35a71a5

+

+=

+

+ 3) 11m10

1m1213m93m11

+

+=

+

+

4) 2y)6y5)(4y(

3y)6y5)(1y(

+=

+5)

xxx

21

22

1=

+

6) 2b8

b42

b53

+=

+

7) 1225

y1y

1yy

=

++

+ 8) 1n13n2

2n2n

1n1n

+

+=

++

+

9) 6m6

)4m(25

2m2

+=

++

+ 10) 22yy5

3y4)1y3(2

+=

5. To solve the problems based on Quadratic EquationExample 1 : If the square of a number is added to 3 times the number, the sum

is 28. Find the number. Solution : Let the number be = x

Square of the number = x2

3 times the number = 3xSquare of a number + 3 times the number = 28

x2 + 3x = 28x2 + 3x 28 = 0

x2 + 7x 4x 28 = 0x(x + 7) 4 (x + 7) = 0(x + 7) (x 4) = 0x + 7 = 0 or x 4 = 0x = 7 or x = 4

The required number is 4 or 7

138

Example 2 : Sum of a number and its reciprocal is 5 51

. Find the number.

Solution : Let the number be = y

Reciprocal of the number = y1

(Number) + (its reciprocal) = 5 51

y + y1

= 526

y1y2 +

= 526

5(y2 + 1) = 26y5y2 + 5 = 26y5y2 26y + 5 = 05y2 25y 1y + 5 = 05y (y 5) 1 (y 5) = 0(y 5) (5y 1) = 0

Either (y 5) = 0 or (5y 1) = 0

y = 5 or y = 51

The required number is 5 or 51

Example 3 : The base of a triangle is 4 cms longer than its altitude. If the area ofthe traingle is 48 sq cms. Find the base and altitude.

Solution : Let the altitude = x cms.Base of the triangle = (x + 4) cms.

Area of the triangle = 21

(base) (height)

48 = 21

(x + 4)x

142

9) A dealer sells an article for Rs. 24 and gains as much percent as the cost priceof the article. Find the Cost price of the article.

10) Sowmya takes 6 days less than the number of days taken by Bhagya to completea piece of work. If both Sowmya and Bhagya together can complete the samework in 4 days. In how many days will Bhagya complete the work?

6. Nature of the roots of a quadratic equation.1) Consider the equation x2 2x + 1 = 0

This is in the form ax2 + bx + c = 0The coefficients are a = 1, b = 2, c = 1

x = a2

ac4bb 2

x = 1x2

1x1.4)2()2( 2 +

x = 2

442

x = 202+

x = 202+

or x = 202

x = 1 or x = 1 roots are equal

2) Consider the equation x2 2x 3 = 0This is in the form ax2 + bx + c = 0the coefficients are a = 1, b = 2, c = 3

x = a2

ac4bb 2

x = 1x2

16)2(

x = 242+

143

x = 242+

or x = 242

x = 26

or x = 22

x = 3 or x = 1 roots are distinct

3) Consider the equation x2 2x + 3 = 0This is in the form ax2 + bx + c = 0The coefficients are a = 1, b = 2, c = 3

x = a2

ac4bb 2

x = 1x2

)3)(1(4)2()2( 22

x = 2

1242

x = 2

82

x = 2

222

x = ( )

2212

= 21

x = 21 + or 21 roots are imaginary

From the above examples it is clear that,1) Nature of the roots of quadratic equation depends upon the value of (b2 4ac)2) The Expression (b2 4ac) is denoted by (delta) which determines the nature

of the roots.3) In the equation ax2 + bx + c = 0 the expression (b2 4ac) is called the discriminant.

144

Discriminant (b2 4ac) Nature of the roots = 0 Roots are real and equal > 0 (Positive) Roots are real and distinct < 0 (negative) Roots are imaginary

Example 1 : Determine the nature of the roots of the equation 2x2 5x 1 = 0.Consider the equation 2x2 5x 1 = 0This is in form of ax2 + bx + c = 0The co-efficient are a = 2, b = 5, c = 1

= b2 4ac = (5)2 4(2) (1) = 25 + 8 = 33

> 0Roots are real and distinct

Example 2 : Determine the nature of the roots of the equation 4x2 4x + 1 = 0Consider the equation 4x2 4x + 1 = 0This is in the form of ax2 + bx + c = 0The co-efficient are a = 4, b = 4, c = 1

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

= 0 Roots are real and equal

Example 3 : For what values of m roots of the equation x2 + mx + 4 = 0 are(i) equal (ii) distinctConsider the equation x2 + mx + 4 = 0This is in the form ax2 + bx + c = 0the co-efficients are a = 1, b = m, c = 4

= b2 4ac = m2 4(1) (4) = m2 16

1) If roots are equal = 0 m2 16 = 0

m2 = 16 m = 16 m = 4

145

2) If roots are distinct > 0 m2 16 > 0 m2 > 16

m2 > 16m > 4

Example 4 : Determine the value of k for which the equation kx2 + 6x + 1 = 0 hasequal roots. Consider the equation kx2 + 6x + 1 = 0 This is in the form ax2 + bx + c = 0 the co-efficients are a = k, b = 6, c = 1

= b2 4ac

since the roots are equal, b2 4ac = 0 ( = 0)(6)2 4(k)(1) = 036 4k = 04k = 36

k = 436

= 9

k = 9Example 5 : Find the value of p for which the equation x2 (p + 2) x + 4 = 0 has

equal roots. Consider the equation x2 (p + 2) x + 4 = 0 This is in the form ax2 + bx + c = 0 Coefficients are a = 1, b = (p + 2), c = 4since the roots are equal = 0

b2 4ac = 0[(p + 2)]2 4(1)(4) = 0(p + 2)2 16 = 0p + 2 = 16p + 2 = 4p + 2 = + 4 or p + 2 = 4 p = 4 2 or p = 4 2 p = 2 or p = 6

146

If m and n are the roots of thequadratic equationax2 + bx + c = 0

Sum of the roots a

b=

Product of roots a

c+=

Exercise : 5.6A. Discuss the nature of roots of the following equations

1) y2 7y + 2 = 0 2) x2 2x + 3 = 0 3) 2n2 + 5n 1 = 04) a2 + 4a + 4 = 0 5) x2 + 3x 4 = 0 6) 3d2 2d + 1 = 0

B. For what positive values of m roots of the following equations are1) equal 2) distinct 3) imaginary1) a2 ma + 1 = 0 2) x2 mx + 9 = 03) r2 (m + 1) r + 4 = 0 4) mk2 3k + 1 = 0

C. Find the value of p for which the quadratic equations have equal roots.1) x2 px + 9 = 0 2) 2a2 + 3a + p = 0 3) pk2 12k + 9 = 04) 2y2 py + 1 = 0 5) (p + 1) n2 + 2(p + 3) n + (p + 8) = 06) (3p + 1)c2 + 2 (p + 1) c + p = 0

7. Relationship between the roots and co-efficient of the terms of the quadraticequation.If m and n are the roots of the quadratic equation ax2 + bx + c = 0 then

m =a2

ac4bb 2 +, n =

a2ac4bb 2

m + n = a2

ac4bb 2 + +

a2ac4bb 2

m + n = a2

ac4bbac4bb 22 +

m + n = a2b2

m + n = a

b-

mn =

+

a2ac4bb 2

a2ac4bb 2

mn = ( )

2

222

a4ac4b)b(

147

mn = ( )

2

22

a4ac4bb

mn = 2

22

a4ac4bb +

mn = 2a4ac4

= a

c mn =

a

c

Example 1 : Find the sum and product of the roots of equation x2 + 2x + 1 = 0x2 + 2x + 1 = 0

This is in the form ax2 + bx + c = 0 The coefficients are a = 1, b = 2, c = 1Let the roots be m and n

i) Sum of the roots m + n = a

b = 1

2

m + n = 2

ii) Product of the roots mn = a

c = 1

1

mn = 1

Example 2 : Find the sum and product of the roots of equation 3x2 + 5 = 03x2 + 0x + 5 = 0

This is in the form ax2 + bx + c = 0 The coefficients are a = 3, b = 0, c = 5Let the roots are p and q

i) Sum of the roots p + q = a

b = 3

0

p + q = 0

ii) Product of the roots pq = a

c = 3

5 pq = 3

5

148

Example 3 : Find the sum and product of the roots of equation 2m2 8m = 02m2 8m + 0 =0

The coefficients are a = 2, b = 8, c = 0Let the roots be and

i) Sum of the rootsa

b=+

2)8(

= = 4

ii) Product of the rootsa

c=

20

= = 0

Example 4 : Find the sum and product of the roots of equation x2 (p+q)x + pq = 0x2 (p + q) x + pq = 0

The coefficients are a = 1, b = (p + q), c = pq

i) Sum of the roots m + n = a

b

m + n = ( )[ ]1

qp +

m + n = (p + q)

ii) Product of the roots mn = a

c = 1

pq

mn = pq

Exercise : 5.7Find the sum and product of the roots of the quadratic equation :1) x2 + 5x + 8 = 0 2) 3a2 10a 5 = 0 3) 8m2 m = 24) 6k2 3 = 0 5) pr2 = r 5 6) x2 + (ab) x + (a + b) = 0

8. To form an equation for the given rootsLet m and n are the roots of the equation

x = m or x = n

i.e., x m = 0, x n = 0(x m) (x n) = 0

x2 mx nx + mn = 0x2 (m + n) x + mn = 0

149

If m and n are the roots then the Standard form of the equation isx2 (Sum of the roots) x + Product of the roots = 0 x2 (m + n) x + mn = 0

Example 1 : Form the quadratic equation whose roots are 2 and 3Let m and n are the rootsm = 2, n = 3Sum of the roots = m + n = 2 + 3

m + n = 5Product of the roots = mn

= (2) (3)

mn = 6Standard form x2 (m + n) x + mn = 0

x2 (5)x + (6) = 0 x2 5x + 6 = 0

Example 2 : Form the quadratic equation whose roots are 52

and 25

Let m and n are the roots

m = 52

and n = 25

Sum of the roots = m + n = 52

+ 25

= 10254+

m + n = 1029

Product of the roots = mn = 25

x52

mn = 1

Standard form x2 (m + n) x + mn = 0

x2 1029

x + 1 = 0

10x2 29x + 10 = 0

150

Example 3 : Form the quadratic equation whose roots are 3 + 2 5 and 3 2 5Let m and n are the roots m = 3 + 2 5 and n = 3 2 5Sum of the roots = m + n

= 3 + 2 5 + 3 2 5 m + n = 6

Product of the roots = mn= (3 + 2 5 ) (3 2 5 )= (3)2 (2 5 )2= 9 20

mn = 11x2 (m + n) x + mn = 0 x2 6x 11 = 0

Example 4 : If m and n are the roots of equation x2 3x + 1 = 0 find the value

of (i) m2n + mn2 (ii) n

1m

1+

Consider the equation x2 3x + 1 = 0This is in the form ax2 + bx + c = 0The coefficients are a = 1, b = 3, c = 1Let m and n are the roots

i) Sum of the roots m + n = a

b = 1

)3( = 3

m + n = 3

ii) Product of the roots mn = a

c

mn = 11

mn = 1

(i) m2n + mn2 = mn (m + n)= 1(3) = 3

(ii)m

1 +

n

1=

mn

mn + =

mn

nm + = 1

3

m

1 +

n

1 = 3

151

Example 5 : If m and n are the roots of equation x2 3x + 4 = 0 form theequation whose roots are m2 and n2. Consider the equation x2 3x + 4 = 0 The coefficients are a = 1, b = 3, c = 4Let m and n are the roots

i) Sum of the roots = m + n = a

b = 1

)3(

m + n = 3

ii) Product of the roots = mn = a

c = 1

4

mn = 4 If the roots are m2 and n2Sum of the roots m2 + n2 = (m + n)2 2mn

= (3)2 2(4)= 9 8

m2 + n2 = 1Product of the roots m2n2 = (mn)2

= 42

m2n2 = 16x2 (m2 + n2) x + m2n2 = 0 x2 (1)x + (16) = 0 x2 x + 16 = 0

Example 6 : If one root of the equation x2 6x + q = 0 is twice the other, find thevalue of q Consider the equation x2 6x + q = 0 This is in the form ax2 + bx + c = 0 The coefficients are a = 1, b = 6, c = qLet the m and n are the roots

i) Sum of the roots m + n = a

b = 1

)6(

m + n = 6

152

ii) Product of the roots mn = a

c = 1

q

mn = qIf one root is (m) then twice the root is (2m)

m = m and n = 2m m + n = 6m + 2m = 6 3m = 6

m = 36

m = 2

We know that q = mnq = m(2m)q = 2m2

q = 2(2)2q = 8

q = 8

Example 7 : Find the value of k so that the equation x2 2x + (k + 3) = 0 has oneroot equal to zero.Consider the equation x2 2x + (k + 3) = 0The coefficients are a = 1, b = 2, c = k + 3Let m and n are the roots

Product of the roots = mn

mn = a

c

mn = 13k +

mn = k + 3Since m and n are the roots, and one root is zero then

m = m and n = 0 mn = k + 3 m(0) = k + 3 0 = k + 3 k = 3

153

Exercise : 5.8

A. Form the equation whose roots are

1) 3 and 5 2) 6 and 5 3) 2 and 23

4) 32

and 23

5) 2 + 3 and 2 3 6) 3 + 2 5 and 3 2 5B.1) If m and n are the roots of the equation x2 6x + 2 = 0 find the value of

i) (m + n) mn ii) m

1 +

n

1

2) If a and b are the roots of the equation 3m2 = 6m + 5 find the value of

i) a

bba+ ii) (a + 2b) (2a + b)

3) If p and q are the roots of the equation 2a2 4a + 1 = 0 Find the value ofi) (p + q)2 + 4pq ii) p3 + q3

4) Form a quadratic equation whose roots are qp

and pq

5) Find the value of k so that the equation x2 + 4x + (k + 2) = 0 has one root equalto zero.

6) Find the value of q so that the equation 2x2 3qx + 5q = 0 has one root whichis twice the other.

7) Find the value of p so that the equation 4x2 8px + 9 = 0 has roots whosedifference is 4.

8) If one root of the equation x2 + px + q = 0 is 3 times the other prove that 3p2 = 16q

Graphical method of solving a Quadratic EquationLet us solve the equation x2 4 = 0 graphically,x2 4 = 0 x2 = 4let y = x2 = 4 y = x2

and y = 4

154

y = x2

x = 0 y = 02 y = 0x = 1 y = 12 y = 1x = 2 y = 22 y = 4x = 1 y = (1)2 y = 1x = 2 y = (2)2 y = 4

x 0 1 1 2 2 3y 0 2 2 8 8 6(x, y) (0, 0) (1, 2) (1, 2) (2, 8) (2, 8) ( 3 ,6)

Step 1: Form table ofcorresponding valuesof x and ySatisfying the equationy = x2

Step 2: Choose the scale onx axis, 1 cm = 1 unity axis, 1 cm = 1 unit.

Step 3: Plot the points (0, 0);(1, 1); (1, 1); (2, 4)and (2, 4) on graphsheet.

Step 4: Join the points by asmooth curve.

Step 5: Draw the straight liney = 4 Parallel to x-axis

Step 6: From the intersectingpoints of the curve andthe line y = 4, drawperpendiculars to thex axis

Step 7: Roots of the equations are x = +2 or x = 2

The graph of a quadratic polynomial is a curve called parabola

Example 1 : Draw a graph of y = 2x2 and find the value of 3 , using the graph.Step 1: Form the table of

corresponding values ofx and y satisfying theequation y = 2x2

Step 2: Choose the scale on xaxis, 1 cm = 1 unit andy axis, 1 cm = 1 unit

Step 3: Plot the points (0, 0);(1, 2) (1, 2); (2, 8) and(2, 8) on graph sheet.

155

x 0 1 1 2 2

y 0 1 1 4 4

(x, y) (0, 0) (1, 1) (1, 1) (2, 4) (2, 4)

x 0 1 1 2 2

y 2 1 3 0 4

(x, y) (0, 2) (1, 1) (1, 3) (2, 0) (2, 4)

Step 4: Join the points by asmooth curve

Step 5: Draw the straight liney = 6 Parallel to x-axis.

Step 6: From the intersectingpoints of the curve andthe line y = 6, drawperpendiculars to thex-axis.

Step 7: Value of 3 = 1.7x = 1.7 or x = + 1.7

Example 2 : Draw a graph of y = x2 and y = 2-x and hence solve the equationx2 + x 2 = 0

Step 1: Form the table ofcorresponding values ofx and y satisfying theequation y = x2

Step 2: Form the table ofcorresponding values ofx and y satisfying theequation y = 2 x.

Step 3: Choose the scale on xaxis 1 cm = 1 unit andy axis, 1 cm = 1 unit.

Step 4: Plot the points (0, 0);(1, 1); (1, 1); (2, 4)and (2, 4) on the graphsheet.

Step 5: Join the points by asmooth curve.

Step 6: Plot the points (0, 2) ;(1, 1); (1, 3); (2, 0)and (2, 4) on graphsheet

156

x 0 1 1 2 2

y 2 1 1 4 4

(x, y) (0, 0) (1, 1) (1, 1) (2, 4) (2, 4)

x 0 1 2 1 2

y 2 3 4 1 0

(x, y) (0, 2) (1, 3) (2, 4) (1, 1) (2, 0)

Step 7: Join the points to get a line.Step 8: From the intersecting

Curve and the line, drawperpendiculars to thex-axis

Step 9: Roots of the equation are x = 1 or x = 2

Example 3 : Solve the equationMethod I : x2 x 2 = 0

Split the equationy = x2 and y = 2 + x

Step 1: Form the table ofcorresponding values xand y satisfying theequation y = x2

Step 2: Form the table ofcorresponding values xand y satisfying theequation y = 2 + x

Step 3: Choose the scale onx axis, 1 cm = 1 unity axis, 1 cm = 1 unit

Step 4: Plot the points (0, 0);(1, 1); (1, 1); (2, 4)and (2, 4) on the graphsheet.

Step 5: Join the points by asmooth curve

Step 6: Plot the points (0, 2);(1, 3) (2, 4); (1, 1) and(2, 0) on the graphsheet.

Step 7: Join the points to get astraight line

Step 8: From the intersectingpoints of Curve and theline, draw the perpendi-culars to the x-axis.

Step 9: Roots of the equation are x = 1 or x = 2

157

x 0 1 1 2 2

y 2 2 0 0 4

(x, y) (0, 2) (1, 2) (1, 0) (2, 0) (2, 4)

Method II :Step 1: Form the table of

corresponding values ofx and y satisfyingequation y = x2 x 2.

Step 2: Choose the scale on xaxis 1 cm = 1 unit andy axis 1 cm = 1 unit.

Step 3: Plot the points (0, 2);(1 2); (1, 0); (2, 0)and (2, 4) on the graphsheet.

Step 4: Join the points to forma smooth curve

Step 5: Mark the intersectingpoints of the curve andthe x axis.

Step 6: Roots of the equations are x = 1 or x = 2

Exercise : 5.9A. 1) Draw the graph of y = x2 and find the value of 7

2) Draw the graph of y = 2x2 and find the value of 3

3) Draw the graph of y = 21

x2 and find the value of 10

B. 1) Draw the graph of y = x2 and y = 2x + 3 and hence solve the equationx2 2x 3 = 0

2) Draw the graph of y = 2x2 and y = 3 x and hence solve the equation2x2 + x 3 = 0

3) Draw the graph of y = 2x2 and y = 3 + x and hence solve the equation2x2 x 3 = 0

C. Solve graphically1) x2 + x 12 = 0 2) x2 5x + 6 = 0 3) x2 + 2x 8 = 04) x2 + x 6 = 0 5) 2x2 3x 5 = 0 6) 2x2 + 3x 5 = 0