01 02 common roots and signs of the quadratic expressions

8/11/2019 01 02 Common Roots and Signs of the Quadratic Expressions

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ii. Common roots and Signs of the Quadratic Expressions

EXERCISE ii

Very Short Answer Questions

1. If the quadratic equations 2 2 0ax bx c+ + = and 2 2 0ax cx b+ + = ( )b c have a

common root then show that a + 4b + 4c = 0

Sol: Let be the common root of 2 2 0ax bx c+ + = and 2 2 0ax cx b+ + =

2 2 0a b c + + =

2 2 0a c b + + =

Subtracting 2 ( ) 0b c c b + =

2 ( ) ( )b c b c =

1

2 =

Sub1

2 = in 2 2 0a b c + + =

1 12 0

4 2a b c

+ + =

a + 4b + 4c = 0

2. If 2 6 5 0x x + = and2

12 0x x p + = have a common root then find p

Sol: Let be the common root of 2 6 5 0x x + = and 2 12 0x x p + =

26 5 0 (1) + =

212 0 (2)p + =

From (1) ( 1) ( 5) 0 =

1 = pr 5=

Substituting 1= in (2) we have

1 12 0 11p p + = =

Substituting 5 = in (2) we have


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25 (2)(5) 0 11p p + = =

3. If2

6 5 0x x + = and2

3 35 0x ax + = have a common root then find a

Sol: Let be the common root of 2 6 5 0x x+ + = and 2 3 35 0x ax + =

26 5 0 (1) + =

23 35 0 (2)a + =

From (1) we have

( 1)( 5) 0 =

1 = or 5 =

If 1 = then

21 3 (1) 35 0 12a a + = =

If 5 = then

25 3 (5) 35 0 4a a + = =

4. If equations 2 0x ax b+ + = and 2 0x cx d+ + = have a common root and the first

equation has equal roots then prove that 2(b + d) = ac

Sol: Let be the common root of 2 0x ax b+ + = and 2 0x cx d+ + = since 2 0x ax b+ + =

has equal roots

, be the roots of 2 0x ax b+ + =

2

aa + = =

2b b = =

is also a root of 2 0x cx d+ + =

2

0c d + + =

02

ab c d

+ + =

2

acb d+ =


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2 ( b +d ) = ac

5. Discuss the signs of the following quadratic expressions when x is real

i) 2 5 4x x +

here2 2

4 ( 5) 4(1) (4)b ac =

= a > 0

Roots are real

Coefficient of2

1 0x = >

Roots of2

5 4 0x x + =

Are 1, 4

By theorem 2

If x < < then 2ax bx c+ + and a have opposite sign here 1 = , 4 =

Coefficient of 2 0x >

21 4 5 4 0x x x < < + <

If x = 1 or x = 4 then

25 4 0x x + =

x < or x < then x < and a have same sign

1x < or 4x < then

25 4 0x x + >

ii)2

3x x +

Here2 2

4 ( 1) 4(1) (3)b ac =

= 1 12 = - 11 < 0

There fore roots of the equation are imaginary coefficient of 2x is 1 > 0

Hence x R 2ax bx c+ + and a have same sign

23 0x x x R + >


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6. For what values of x the following expressions are positive

i)2

5 6x x + ii)2

3 4 4x x + iii)2

4 5 2x x + iv)2

5 14x x +

Sol:2

5 6 0x x + = Have real roots 2 and 3

When x < 2 or x > 32

5 6 0x x + >

25 6 0x x + > is possible when ( , 2) (3, )x

ii) 23 4 4x x+ +

24 16 (3)(4) 32 0b ac = = = <

Roots of 23 4 4 0x x+ + = are imaginary

Hence 23 4 4 0x x+ + > x R because coefficient of 2 3 0x = >

iii)2

4 5 2x x +

2 24 (4) 4( 5) (2)b ac = =

= 16 + 40 =56

Roots of 24 5 2 0x x + =

25 4 2 0x x =

4 16 4(5)( 2)

10x

=

4 56

10x

=

If4 56 4 56

10 10x

+<

Then2

4 5 2 0x x + >

2x ax bx c < < + + and a have opposite sign

iv) 2 5 14x x +

Here2

4b ac =

= 25 4(1) (14) = -31 < 0


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Roots of 2 5 14 0x x= + =

Are imaginary

25 14 0x x x R + >

7. For what values of x the following expressions are negative

i) 2 7 10x x + ii) 215 4 3x x+ + iii) 22 5 3x x+ iv) 2 5 6x x

i) 2 7 10x x +

Here 2 4 49 40 9b ac = = =

Roots are real coefficient of2

1 0x = >

Roots of the equation2

7 10 0x x + = are 2, 5

If x < < then 2ax bx c+ + and a have opposite sign

22 5 7 10 0x x x < < + <

ii) 215 4 3x x+

a = -3, b = 4, c = 15

24 16 4( 3)(15)b ac = = =

= 196 > 0

Roots are real roots of2

15 4 3 0x x+ =

23 4 15 0x x =

23 9 5 15 0x x x + =

x = -5/3, x = 3

Coefficient of2

3 0x = <

215 4 3 0x x + < When

5

3x < and x > 3

x < or 2x ax bx c> + + and a have same sign


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iii) 22 5 3x x+

a = 2, b = 5, c = - 3

24 25 24 49b ac = = + =

Roots are real

2 22 5 3 0 2 6 3 0x x x x x+ = + =

(2x-1)(x + 3) = 0

Roots of2

2 5 3 0x x+ =

Are1

3,2

Coefficient of 2 2 0x = >

22 5 3 0x x + < When

13

2x < <


iv) 2 5 6x x

a = 1, b = - 5, c = -6

24 25 24 49b ac = = + =

Roots are real

25 6 0 ( 6)( 1) 0x x x x = + =

x = 6; x = -1

Coefficient of 2 1x =

25 6 0 6 1x x x < < <



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8. Find the changes in sign of the following expressions and find their extreme values

i)2

5 6x x +

Here a = 1, b = - 5, c = 6

24 25 24 1b ac = = =

Hence roots are real

25 6 0 2,3x x x + = =

22 3 5 6 0x x x < < + <

22 3 5 6 0x or x x x= = + =

22 3 5 6 0x or x x x< > + >

Here a = 1 > 0

25 6x x + has absolute minimum value

absolute minimum value2

4 24 25 1

4 4 4

ac b

a

= = =

ii) 215 4 3x x+

a = -3, b =4, c = 15

24 16 4( 3)(15)b ac = =

= 196 > 0

Roots are real roots of2

15 4 3 0x x+ = are5

3 and 3

{Refer to prob (ii) of 7}

25 3 15 4 3 0

3

x x x < < + >

25 3 15 4 3 03

x or x x x< > + <

253 15 4 3 0

3x or x x x< > + =


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Here a = - 3 < 0

Absolute maximum value

24 180 16

4 4 3

ac b

a x

= =

196 4912 3

= =

9. Find the maximum or minimum values of following expressions as x varies over R

i)2

7x x +

Here a = 1, b = -1, c = 7

a > 0 absolute minimum2

4

4

ac b

a

=

28 1 27

4 4

= =

ii) 212 32x x

a = -1, b = 12, c = -32

a < 0 hence absolute maximum value2

4 49 1)( 32) 144

4 4

ac b

a

= =

= 4

iii) 22 5 3x x+

Here a = -3, b = 2, c = 5

a < 0 hence maximum value2

4

4

ac b

a

=

60 4 16

12 3

= = +

iv)

2

ax bx a+ +

( , )a b R

a > 0 minimum value2 2

4

4

a b

a

=

a < 0 maximum value2 2

4

4

a b

a

=


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SHORT ANSWER QUESTIONS:

Determine the range of the following expressions

1.2

2

1

1

x x

x x

+ +

+

Sol: let2

2

1

1

x xy

x x

+ +=

+

2 21x x x y xy y + + = +

2( 1) ( 1) 1 0x y x y y + + =

x is real

24 0b ac

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

23 10 3 0y y +

23 10 3 0y y +

1(3 1)( 3) 0 3

3y y y

ii) 22

2 3 6

x

x x

+

+ +

Let2

2

2 3 6

xy

x x

+=

+ +

22 2 3 6x x y xy y+ = + +

22 (3 1) 6 2 0x y x y y+ + =

Since x is real 2 4 0b ac

2(3 1) 4(2 )(6 2) 0y y y

2 2(9 6 1) (48 16 ) 0y y y y +

239 10 1 0y y + +

239 10 1 0y y


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13 (3 1) 1(3 1) 0y y y +

(13 1)(3 1) 0y y+

1 1

13 3y

iii) ( 1)( 2)3

x xx

+

+

Let( 1)( 2)

3

x xy

x

+=

+

22 3x x xy y + = +

2(1 ) (2 3 ) 0x x y y+ + =

Since x is real

24 0b ac

2(1 ) 4(2 3 ) 0y y + +

22 1 8 12 0y y y + + +

210 7 0y y +

( 1)( 9) 0y y

1y or 9y

( , 1] [9, )y

iv)2

2

2 6 5

3 2

x x

x x

+

+

Let2

2

2 6 5

3 2

x xy

x x

+=

+

2 22 6 5 3 2x x x y xy y + = +

2(2 ) 6 3 5 2 0x y x xy y + + =

2(2 ) 3 ( 2) 5 2 0x y x y y + + =

Since x is real2

4 0b ac


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29( 2) 4(2 )(5 2 ) 0y y y

2 29 36 36 4{10 9 2 } 0y y y y + +

24 0 ( 2)( 2) 0y y y +

( , 2] [2, )y

2. Prove that1 1 1

3 1 1 (3 1)( 1)x x x x+

+ + + +does not lie between 1 and 4 if x is real

Sol: Let1 1 1

3 1 1 (3 1)( 1)y

x x x x+ =

+ + + +

1 3 1 1

(3 1)( 1)

x xy

x x

+ + + =

+ +

24 1

3 4 1x y

x x+ =

+ +

24 1 3 4x x y xy y+ = + +

23 4 ( 1) 1 0x y x y y+ + =

Since x is real

24 0b ac

2{4( 1)} 4(3)( )( 1) 0y y y

2 216 32 16 12 12 0y y y y + +

24 20 16 0y y +

25 4 0y y +

( 1)( 4) 0y y

( ,1] [4, )y

Hence y does not lie between 1 and 4


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3. If x is real then prove that2 5 9

x

x x +lies between

1

11 and 1

Sol: Let2 5 9

xy

x x=

+

25 9x x y xy y= +

2(5 1) 9 0x y y x y + + =

Since x is real

24 0b ac

2(5 1) 4 (9 ) 0y y y+

2 225 10 1 36 0y y y+ +

211 10 1 0y y + +

211 10 1 0y y

211 11 1 0y y y +

(11 1)( 1) 0y y+

11

11

y

Hence y lies between1

11 and 1

4. If the expression2

3 2

x p

x x

+takes all real values x R then find the bounds of p

Sol: Let2

3 2

x py

x x

=

+

2

3 2x p x y xy y = +

2(3 1) 2 0x y x y y p + + + =

Since x is real

24 0b ac


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2(3 1) 4 (2 ) 0y y y p+ +

26 4 1 0y y py+ +

2(6 4 ) 1 0y y p+ +

Since coefficient of2

y and expression have same sign y R

2(6 4 ) 4 0p

236 16 48 4 0p p+

216 48 32 0p p +

23 2 0p p +

( 1)( 2) 0p p

(1, 2)p

{ for p 1, 2, given expression is not defined}

5. If 2c ab and the roots of2 2 2 2

( ) 2( ) ( ) 0c ab x a bc x b ac + = are equal then show

that3 3 3

3a b c abc+ + = (or) a = 0

Sol: Given equation is

2 2 2 2

( ) 2( ) ( ) 0c ab x a bc x b ac + =

Roots are equal

2 2 2 24( ) 4( )( ) 0a bc c ab b ac =

4 2 2 2 2 2 3 3 22 0a b c a bc b c ac ab a bc+ + + =

4 3 3 23 0a ab ac a bc+ + =

3 3 3( 3 ) 0a a b c abc+ + =

0a = or3 2 3

3 0a b c abc+ + =


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III) QUADRATIC INEQUATIONS

EXERCISE -- 1(c)

1. Solve the in equations by algebraic method

i) 215 4 4 0x x+

215 10 6 4 0x x x+

5 (3 2) 2(3 2) 0x x x+ + <

(3 2)(5 1) 0x x+

2 / 3 2 / 5x

ii) 2 2 1 0x x + <

2( 1) 0x <

Square of any real number cannot be negative

Hence x

iii)2

2 3 2 0x x

22 3 2 0x x+

22 4 2 0x x x+

2 ( 2) ( 2) 0x x x+ +

( 2)(2 1) 0x x+

12

2x

iv)2

4 21 0x x

2

7 3 21 0x x x +

( 7) 3( 7) 0x x x +

( 3)( 7) 0x x+

3x or 7x

( , 3] [7, )x


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2. Solve the following in equations by graphical method

i) 2 7 6 0x x + >

let2

7 6y x x= +

The values of y for some selected values of x are given below

x -5 -4 -3 -2 -1 0 1

y 66 50 36 24 14 6 0

x 2 3 4 5 6

y -8 -6 -6 -4 0

Hence solution set ( , 1] [6, )=

ii) 24 0x

Let2

4y x=

x -4 -3 -2 -1 0 1 2 3 4

y -12 -5 0 3 4 3 0 -5 -12

27 6 0y x x= + 1 6when x or x

5 4 3 2 1 1 2 3 4 5

24 0y x= 2 2when x

5 4 3 2 1 1 2 3 4 5


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Hence solution set [ 2, 2]x

iii) 215 4 4 0x x+

let2

15 4 4y x x= +

x -3 -2 -1 2

3

0 2

5

1

y 89 48 7 0 -4 0 15

215 4 4 0y x x= +

2 2

3 5x

Hence2 2

,3 5

x

iv) 2 4 21 0x x

24 21y x x=

x -3 -2 -1 0 1 2 7

y 0 -9 -16 -21 -24 -25 0

2

52

5


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24 21 0y x x=

When 3x or 7x

2. Solve the following in equations

3 8 2x <

83 8 0 (1)

3x x

And 3 8 2x <

Here 3 8 0x

But given 3 8 2x <

No solution

ii) 2 6 5 8 2x x x + >

2 6 5 0x x +

And i) 8 2x < 0 or2 2

6 5 (8 2 )x x x + > and 8 2 0x

We have 2 6 5 0x x +

( 1)( 5) 0 1 5x x x

And 2 8 0 4x x > >

73

0


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Or 2 26 5 64 4 32x x x x + > +

20 5 38 69x x> +

25 38 69 0x x + <

And x < 4

25 23 15 69 0x x x + <

(5 23) 3(5 23) 0x x x <

( 3)(5 23) 0x x <

233

5x< <

1 5x and x > 4

Or23

35

x< < and x < 4

3 5x <

iii) Solve the in equation2

2 8x x+ >

When a, b R and 0a , 0b then 0a b a b> >

22 8x x + >

22 (8 ) 0x x + >

2, 2 2x x> <

2 6 0x x+ >

( , 3) (2, )x

Solution

(2, 2 2)x


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iv) Solve the in equation

2( 3)(2 ) 4 12 11x x x x < + +

( 3)(2 ) 0x x and2

( 3)(2 ) 4 12 11x x x x < + +

2 3x and2 2

5 6 4 12 11x x x x < + +

2 3x and 20 5 7 17x x< + +

2 3x and 25 7 17 0x x+ +

25 7 17x x o+ + > x R

27 4(5)(17) 0 <

Solution set [2, 3]

v) Solve the in equation

2 26 6

2 5 4

x x x x

x x

+ +

+ +

26 0x x+ = , 2 5 0x + , and 4 0x +

(or)2

6 0x x+ > and1 1

2 5 4x x

+ +

26 0 2x x x+ = or x = 3

26 0 2 3x x x+ > < <

1 12 5 4

2 5 4x x

x x + +

+ +

1x

{ }2,3 [ 2, 1]

[ 2, 1] {3}


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vi) Solve the in equation

23 10 8x x x >

Sol: 2 3 10 0x x

and i) 8 x < 0

or (ii)2 2

3 10 (8 )x x x > and (8 ) 0x

( 5) ( 2) 0x x +

[ , 2] [5, ]x

8 0 8x x < >

ii) 2 23 10 64 16x x x x > +

7413 74

13x x> >

And 8 0 8x x

74,

13x

( ,8)x

74(8, ) , ( ,8)13

74 74(8, ) , ,

13 13

=

SOLVE THE FOLLOWING EQUATIONS

1. 22

1 19 27 8 0x x

x x

+ + + =

Let1

x ax

+ =

2 2 2 2

2 2

1 12 2x a x a

x x+ + = + =


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2

2

1 19 27 8 0x x

x x

+ + + =

29( 2) 27 8 0a a + =

29 27 10 0a a =

29 3 30 10 0a a a+ =

3 (3 1) 10(3 1) 0a a a+ + =

(3 1)(3 10) 0a a+ =

1

3a = ,

10

3a =

Case i): suppose1

3

a =

21 1 1 1

3 3

xx

x x

++ = =

23 3 3 3 0x x x x2 + = + + =

1 1 4(3)(3)

2 3x

=

1 35

6

i

x

=

Case ii):10

3a =

21 10 3 3 103

x x xx

+ = + =

23 10 3 0x x + =

2

3 9 3 0x x x + =

3 ( 3) 1( 3) 0x x x =

(x 3) (3x -1) = 0

3x = ,1

3x=


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23/26

1

3a = , a = 3

Case i)1

3a =

4 1 1

36 9 4 14 1 3

x

x xx

= = ++

10 532 10

32 16x x= = =

Case ii) a = 3

4 1 4 13 9

4 1 4 1

x x

x x

= =

+ +

4 1 36 9 32 10x x x = + =

5

16x= is only solution

5

16x= Does not satisfy the equation

3. 22

43 1 5

3 1x

x+ + =

+

Let 23 1x a+ =

24 5 5 4 0a a aa

+ = + =

( 4)( 1) 0 4; 1a a a a = = =

Case i) a = 4

2 23 1 4 3 1 16x x+ = + =

2 5 5x x = =

Case ii) a = 1

2 23 1 1 3 1 1 0x x x+ = + = =


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5. 22

1 12 3 1x x

x x

+ + =

This is same model as (1) try your self

6. x (x + 2 ) (x + 3) (x + 5) = 72

X (x +5) (x + 2) (x +3) = 72

2 2( 5 )( 5 6) 72x x x x+ + + =

Let 2 5x x a+ =

2( 6) 72 6 72 0a a a a+ = + =

(a +120 (a 6) = 0

a = - 12, a = 6

case i)a = - 12

2 25 12 5 12 0x x x x+ = + + =

5 25 48 5 23

2 2

ix

= = =

Case ii)a = 6

2 25 6 5 6 0x x x x+ = + =

26 6 0x x x+ =

( 6)( 1) 0x x+ =

6x = , x = 1

7. x (x -1) (x + 2) (x 3) = - 8

2 2( ){ 6} 8x x x x =

Let2

x x a =

2( 6) 8 6 8 0a a a a = + =

(a 4) (a 2) = 0

a 4 or a = 2


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Case i): a = 4

2 24 4 0x x x x = =

1 1 16 1 17

2 2x

+ = =

Case ii) a = 2

2 22 2 0x x x x = =

22 2 0x x x + =

X (x 2) + (x 2) = 0

(x + 1)( x 2) = 0

X = 2; x = -1

8. (x 1) (x + 1) (2x +3) (2x 1) = 3

(x 1) (2x + 3) (x +1) 2x 1) = 3

2 2(2 3 2 3}{2 2 1} 3x x x x x x+ + =

2 2(2 3)(2 1) 3x x x x+ + =

22x x a+ =

2( 3)( 1) 3 4 3 3a a a a = + =

24 0 0,4a a a = =

If 2 22 2 0x x a x x+ = + =

1(2 1) 0 0;

2x x x+ = =

If2 2

2 4 2 4 0x x x x+ = + =

1 1 4(2)( 4) 1 33

4 4x

= =


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9. Find the range of the values of x which satisfy( 1)( 3)

02

x x

x

+

Sol:( 1)( 3)

02

x x

x

+

Case i) ( 1)( 3) 0x x+

And (x 2) > 0

3x

Case ii) ( 1)( 3) 0x x+ and x 2 < 0

1 3x and x < 2

1 2x <

solution set [ 1, 2) [3, )

01 02 common roots and signs of the quadratic expressions

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