Unit # 2 Theory of Quadratic Equations

8/22/2019 Unit # 2 Theory of Quadratic Equations

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Mudassar Nazar Notes Published by Asghar Ali Page 1

Unit # 2 Theory of Quadratic Equations

Exercise # 2.1

Question # 1

Find the discriminent of the following given quadratic equations:

(i) 9x2 30x + 25 = 0

Solution:

9x2

30x + 25 = 0

Here,

a = 9 , b = -30, c = 25

Disc = b2 4ac

Disc = (-30)2 4(9)(25)

Disc = 900 900

Disc = 0

(iv) 4x2 7x 2 = 0

Solution:

4x2 7x 2 = 0

Here,

a = 4, b = -7 , c = -2

Disc = b2

4ac

Disc = (-7)2 4(4)(-2)

Disc = 49 + 32

Disc = 81


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Question # 2

Find the nature of the roots of the following given quadratic equations and verify the result by solving

the equations.

(iii) 16x2 24x + 9 = 0

Solution:

16x2 24x + 9 = 0

Here,

a = 16, b = -24 , c = 9

Disc = b2 4ac

Disc = (-24)2 4(16)(9)

Disc = 576 576

Disc = 0

Hence,

Roots are real rational and equal

Solution of the equation

16x2 24x + 9 = 0

16x2 12x 12x + 9 = 0

4x(4x 3) 3(4x 3) = 0

(4x 3) (4x 3) = 0

(4x 3)2 = 0

4x 3 = 0

4x = 0 + 3

4x = 3

x =


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(iv) 3x2

+ 7x 13 = 0

Solution:

3x2

+ 7x

13 = 0

Here,

a = 3 , b = 7 , c = -13

Disc = b2 4ac

Disc = (-7)2 4(3)(-13)

Disc = 49 + 156

Disc = 205 > 0

Hence,

Roots are real irrational and unequal

Solution of the equation

3x2 + 7x 13 = 0

Here,

a = 3 , b = 7 , c = -13

By quadratic formula

x =

x =

x =

x =


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Question # 4

Find the value of k , if the roots of the following equations are equal:

(ii) x

2

+ 2(k + 2) x + (3k + 4) = 0

Solution:

x2 + 2(k + 2) x + (3k + 4) = 0

Here,

a = 1, b = 2(k + 2) , c = 3k + 4

The roots of the equation will be equal if

Disc = 0

[2(k + 2)]2 4(1)(3k + 4) = 0

4(k2

+ 4k + 4) 12k 16 = 0

4k2 + 16k + 16 12k 16 = 0

4k2 + 4k = 0

4(k2 + k) = 0

K2 + k =

K2 + k = 0

K(k + 1) = 0

k = 0 or k + 1 = 0

k = 0 or k = 0 1

k = 0 or k = -1

(iii) (3k + 2)x2 5(k +1) x + (2k + 3) = 0

Solution:

(3k + 2)x2 5(k +1) x + (2k + 3) = 0


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Here,

a = 3k + 2 , b = -5(k + 1) , c = 2k + 3

The roots of the equation will be equal if

Disc = 0

[-5(k + 1)]2 4(3k + 2) (2k + 3) = 0

[25(k2

+ 2k + 1)] 4[6k2

+ 9k + 4k + 6] = 0

25k2 + 50k + 25 24k2 -36k -16k 24 = 0

K2 + 50k 52k + 1 = 0

K2 2k + 1 = 0

K2

k

k + 1 = 0

K(k 1) k (k 1) = 0

(k 1) ( k 1 ) = 0

(k 1)2 = 0

k 1 = 0

k = 0 + 1

k = 1

Question # 8

Show that the roots of the following equations are rational:

(ii) (a + 2b)x2

+ 2(a + b + c)x + (a + 2c) = 0

Solution:

(a + 2b)x2 + 2(a + b + c)x + (a + 2c) = 0

Disc = [2(a + b + c)]2 4 (a + 2b) (a + 2c)

Disc =[4(a2 + b2 + c2 + 2ab + 2bc + 2ca)] 4[a2 + 2ca + 2ab + 4bc]


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Disc = 4[a2

+ b2

+ c2

+ 2ab + 2bc + 2ca a2 2ca 2ab 4bc]

Disc = 4[b2 + c2 2bc]

Disc = (2)2

(b c)2

Disc = [2(b

c)]2 (perfect square)

Hence,

The roots of the given equation are equal


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Exercise # 2.2

Question # 1

Find the cube roots of -1 , 8, -27, 64

Solution:

Let x be the cube roots of -1

x = (-1)1/3

x3

= -1

x3

+ 1 = 0

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

If x + 1 = 0

x = 0 1

x = -1

If x2 x + 1 = 0

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

By Quadratic Formula

x = * =

x = * = 2

x =

x =

x = or x =

Hence,

Cube roots of -1 are -1 , ,2


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(ii) 8

Let x be the cube roots of 8

x = (8)1/3

x3 = 8

x3 8 = 0

(x)3 (2)

3= 0

(x 2) [(x)2 + (x) (2) + (2)2] = 0

(x 2) (x2 + 2x + 4) = 0

If x 2 = 0

x = 0 + 2

x = 2

If x2 + 2x + 4 = 0

Here,

a = 1 , b = 2 , c = 4


x =

x =

x =

x =

x =

x =

x =

x = or x =


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x = 2 or x = 22

Here,

Cube roots of 8 are 2, 2 , 22

(iii) -27

Let x be the cube roots of -27

x = (-27)1/3

x3 = -27

x3 + 27 = 0

x3 + (3)3 = 0

(x + 3) [(x)2

(x) (3) + (3)2] = 0

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

If x + 3 = 0

x = 0 3

x = -3

If x2 3x + 9 = 0

Here,

a = 1 , b = -3 , c = 9


x =

x =

x =

x =

x =

x =


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x =

x = or x =

x = 3(- ) or x = 3(2)

x = -3 or x = -3 2

Here,

Cube roots of -27 are 3, -3 , -3 2

(iV) 64

Let x be the cube roots of 64

x = (64)1/3

x3 = 64

x3 - 64 = 0

x3 + (4)3 = 0

(x + 4) [(x)2 +(x) (4) + (4)2] = 0

(x + 4) (x2

+ 4x + 16)= 0

If x - 4 = 0

x = 0 + 4

x = 4

If x2

+4x + 16 = 0

Here,

a = 1 , b = 4 , c = 16


x =

x =

x =


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x =

x =

x =

x =

x = or x =

x = 4 or x = 42

Here,

Cube roots of 64 are 4, 4 , 4 2


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Exercise # 2.3

Question # 1

Without solving find the sum and product of the roots of the following quadratic equations:

(iii) px2 qx + r = 0

Solution:

px2 qx + r = 0

Here,

a = p , b = -q , c = r

Sum of roots =

=

=

Product of roots =

=

(iv) (a + b)x2 ax + b = 0

Solution:

(a + b)x2 ax + b = 0

Sum of roots =

=

Product of roots =

(v) (l + m)x2

+ ( m + n)x + n = 0

Solution:

(l + m)x2 + ( m + n)x + n = 0


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Sum of roots =

=

Product of roots =

=

(vi) 7x2- 5mx + 9n = 0

Solution:

7x2 + 5mx + 9n = 0

Sum of roots =

=

=

Product of roots =

=

Question # 2

Find the value of k, If :

(ii) Sum of roots of the equation x2

+ (3k 7) + 5k = 0 is times the product of the roots.

Solution:

x2 + (3k 7) + 5k = 0

Here,

a = 1, b = 3k 7 , c = 5k

Sum of roots =

=

= -3k + 7


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Product of roots =

=

= 5k

As given,

Sum of the roots = (product of the roots)

-3k + 7 = (5k)

2(-3k + 7) = 3(5k)

-6k + 14 = 15k

-6k

15k = -14

-21k = -14

21k = 14

k =

k =

Question # 3

Find the value of k, If :

(ii) Sum of the squares of the roots of the equation x2 2kx + (2k = 1) = 0 is 6

Solution:

x2 2kx + (2k = 1) = 0

Here,

a = 1, b = -2k , c = 2k + 1

Let and be the roots of the equation

Sum of roots =

+ =


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+ = 2k

Product of the roots =

=

= 2k + 1

Given,

2 + 2 = 6

( + )2 2 = 6

(2k)2 2(2k +1) = 6

4k2 4k 2 = 6

4k2 4k 2 6 = 0

4k2 4k 8 = 0

4(k2 k 2) = 0

k2 k 2 =

k2 k 2 = 0

k2 + k 2k 2 = 0

k (k + 1) 2( k + 1) = 0

(k 2) (k + 1) = 0

k 2 = 0 or k + 1 = 0

k = 0 + 2 or k = 0 -1

k = 2 or k = -1

k = 2, -1

Question # 4

Find the value of P, If :

(ii) The roots of equation x2 3x + p 2 = 0 differ by 2


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Solution:

x2 3x + p 2 = 0

Here,

a = 1 , b = -3 , c = p 2

Let and ( - 2) be the roots of the equation

Sum of roots =

+ - 2 =

2 - 2 = 3

2 = 3 + 2

2 = 5

=


( - 2) =

( - 2) = p 2

( ) = p 2

( ) = p 2

= p 2

5 = 4( p 2)

5 = 4p 8

5 + 8 = 4p

13 = 4p

= p

P =


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Question # 5

Find the value of m, If :

(ii) The roots of equation x

2

+ 7x + 3m

5 = 0 , Satisfy the relation 3 - 2 = 4Solution:

x2 + 7x + 3m 5 = 0

Here,

a = 1 , b = 7 , c = 3m 5


Sum of roots =

+ =

+ = -7

= -7 (i)


=

= 3m 5 (ii)

Given,

3 - 2 = 4

From (i)

3(-7 ) - 2 = 4

-21

3 - 2 = 4

-5 = 4 + 21

-5 = 25

=


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= -5

Putting in (i)

= -7(-5)

= - 7 + 5

= - 2

Putting the value of and in (ii)

(-2) (-5) = 3m 5

10 = 3m 5

10 + 5 = 3m

15 = 3m

= m

5 = m

m = 5

(iii) The roots of the equation 3x2 2x + 7m + 2 = 0, Satisfy the relation 7 - 3 = 18

Solution:

3x2

2x + 7m + 2 = 0

Here,

a = 3 , b = -2 , c = 7m + 2


Sum of roots =

+ =

+ =

= (i)



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=

= (ii)

Given,

7 - 3 = 18

From (i)

7( - ) - 3 = 18

- 7 - 3 = 18

-10 = 18 -

-10 =

-10 =

=

=

Putting in (i)

=

( )

= +

=

=

= 2

Putting the value of and in (ii)

(2) ( ) =

=

-8 = 7m + 2


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-8 2 = 7m

-10 = 7m

= m

m =

Question # 6

Find m, If sum and product of the roots of the following equation is equal to a given number

(i) (2m + 3)x + (7m 5)x + (3m 10 ) = 0

Solution:

(2m + 3)x + (7m 5)x + (3m 10 ) = 0

Here,

a = 2m + 3 , b = 7m - 5 , c = 3m 10


Sum of roots =

+ =


=

Given,

= (i)

and

= (ii)

Comparing (i) & (ii)

=


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-7m + 5 = 3m 10

-7m 3m = -10 5

-10m = -15

10m = 15

m =

m =

(ii) 4x2 (3 + 5m)x (9m 17) = 0

Solution:

4x2

(3 + 5m)x

(9m

17) = 0

Here,

a = 4 , b = -(3 + 5m) , c = - ( 9m 17)


Sum of roots =

+ =

+ =


=

=

Given,

= (i)

and

= (ii)


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Comparing (i) & (ii)

=

3 + 5m = -9m + 17

5m + 9m = 17 - 3

14m = 14

m =

m = 1


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Exercise # 2.5

Question # 1

Write the quadratic equation having following roots:

(d) 0 , - 3

Solution:

S = Sum of roots

S = 0 + (-3)

S = 0

3

S = -3

P = Product of roots

P = (0) (-3)

P = 0

The required equation is

x2

Sx + P = 0

x2(-3)x + 0 = 0

x2

+ 3x = 0

(e) 2 , -6

Solution:

S = Sum of roots

S = 2 + (-6)

S = 2 6

S = -4


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P = (2) (-6)

P = -12


x2 Sx + P = 0

x2(-4)x +(-12) = 0

x2 + 4x - 12 = 0

(f) -1 , -7

Solution:

S = Sum of roots

S = (-1) + (-7)

S = -1 7

S = -8


P = (-1) (-7)

P = 7


x2 Sx + P = 0

x2(-8)x +(7) = 0

x

2

+ 8x + 7 = 0

(g) 1 + , 1


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Solution:

S = Sum of roots

S = 1+ + 1 -

S = 2


P = (1+ ) (1 )

P = (1)2 ( )

2

P = 1 (-1)

P = 1 + 1

P = 2


x2 Sx + P = 0

x2(2)x +(2) = 0

x2

- 2x +2 = 0

(h) 3 + , 3 -

Solution:

S = Sum of roots

S = 3 + + 3 -

S = 6


P = (3 + ) (3 - )

P = (3 )2 ( )2


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P = 9 2

P = 7


x2

Sx + P = 0

x2(6)x +(7) = 0

x2

- 6x +7 = 0

Question # 2

If and are the roots of the equation x2 3x + 6 = 0 , Form equation whose roots are:

(d) +

Solution:

x2 3x + 6 = 0

Here,

a = 1 , b = -3 , c = 6

and are the roots of the equation

Sum of the roots =

+ =

+ = 3


=

= 6

Given roots are and

S= +


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S =

S =

S =

S =

S =

S =

P = ( ) ( )

P = 1


x2 Sx + P = 0

x2 ( )x + 1 = 0

x2

+ x + 1 = 0

2x2

+x +2 = 0

(e) + , +

Solution:

Given roots are + and +

S = + + +

S = + +

S = ( + ) +

S = 3 +

S = 3 +


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S =

S =

P =( + ) ( + )

P = ( + ) ( )

P = (3) ( )

P =

Required quadratic equation is

x2 Sx + P = 0

x2 x + = 0

2x2 7x + 3 = 0

Question # 3

If and are the roots of the equation x2 + px + q = 0 .Form equations whose roots are:

(i)2

,2

Solution:

x2

+ px + q = 0

Here,

a = 1 , b = p , c = q

Sum of roots =

+ =

+ = -p

Product of roots =


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=

= q

Given roots are2

,2

S = 2 + 2

S = ( + )2 - 2

S = (-p)2 2q

S = p2 2q

P = (2

) (2)

P = 2 2

P = ( )2

P = q2


x2 - Sx + P = 0

x2(p2 2q)x + q

2 = 0

(ii) ,

Solution:

Given roots are and

S = +

S =

S =

S =


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S =

P = ( ) ( )

P = 1


x2

- Sx + P = 0

x2 ( )x + 1 = 0

Or

qx2 (p2 2q)x + q = 0

Exercise # 2.6


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Question # 1

Use Synthetic Division to find the quotient and remainder, when:

(iii) (x3

+ x2 3x + 2) (x 2)

Solution:

(x3 + x2 3x + 2) (x 2)

1 1 -3 2

2 2 6 6

1 3 3 8

Quotient = Q(x) = x2 + 3x + 3

Remainder= 8

Question # 2

Find the value of h using Synthetic Division, If

(ii) 1 is zero of polynomial x3 2hx

2+ 11

Solution:

P(x) = x3 2hx2 + 11

1 -2h 0 11

1 1 1-2h 1-2h

1 1-2h 1-2h 12-2h

If 1 is the zero of the polynomial , then

R = 0

12-2h = 0


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-2h = 0 12

-2h = -12

2h = 12

h =

h = 6

(iii) -1 is zero of polynomial 2x3

+ 5hx 23

Solution:

P(x) = 2x3

+ 5hx 23

2 0 5h -23

-1 -2 2 -5h-2

2 -2 5h+2 -5h-25

If -1 is the zero of the polynomial , then

R = 0

-5h-25 = 0

-5h = 0 + 25

-5h = 25

h =

h = -5

Question # 3


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Use Synthetic Division to find the value of l and m , If :

(ii) (x -1) and (x + 1) are the factors of the polynomial x3 3lx

2+ 2mx + 6

Solution:P(x) = x3 3lx

2 + 2mx + 6

By Synthetic Division

1 -3l 2m 6

1 1 1- 3l 2m-3l+1

1 1-3l 2m-3l+1 2m-3l+7

If (x -1) is a factor of the polynomial , then

R = 0

2m-3l+7 = 0 (i)

And

1 -3l 2m 6

1 -1 1+3l -2m-3l-1

1 -1-3l 2m+3l+1 -2m-3l+5

If (x -1) is a factor of the polynomial , then

R = 0

-2m-3l+5= 0

-(2m+3l-5) = 0

2m + 3l -5 = 0 (ii)

(i) + (ii)2m -3l + 7 = 0

2m +3l -5 = 0

4m +2 = 0


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4m = 0 2

4m = -2

m =

m =

Putting in (i)

2( ) 3l + 7 = 0

-1 -3l + 7 = 0

-3l+6 = 0

-3l= 0 6

-3l = -6

3l= 6

l =

l = 2

Question # 4Solve by using Synthetic Division, if :

(ii) 3 is the root of the polynomial 2x3

-3x2 11x + 6 = 0

Solution:

P(x) =2x3 -3x2 11x + 6 = 0

By Synthetic Division

2 -3 -11 6

3 6 9 6

2 3 -2 0


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Depressed Equation

2x2 + 3x -2 = 0

2x2 x + 4x -2 = 0

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

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

2x -1 = 0 or x+ 2 = 0

2x=0 +1 or x = 0 2

2x = 1 or x = -2

x = or x = -2

The roots of the equation are -2, , 3

(iii) -1 is the root of the equation 4x3 x

2 11x 6 = 0

Solution:

P(x) = 4x3 x2 11x 6 = 0

4 -1 -11 -6

-1 -4 5 6

4 -5 -6 0

Depressed Equation

4x2 5x 6 = 0

4x2 + 3x

8x

6 = 0

x(4x + 3) -2(4x + 3) = 0

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

x-2 = 0 or 4x + 3 = 0


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x = 0 + 2 or 4x = 0 3

x = 2 or 4x = -3

or x =

Roots of the equations are , -1 , 2

Question # 5

Solve by using Synthetic Division, if :

(ii) 3 and -4 are the roots of the equation x4

+ 2x3 13x

2 14x + 24 = 0

Solution:

x4 + 2x3 13x2 14x + 24 = 0

1 2 -13 -14 24

3 3 15 6 -24

1 5 2 -8 0

4 -4 -4 8

1 1 -2 0

Depressed Equation

x2 + x 2 = 0

x2 x + 2x -2 = 0

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

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

x + 2 = 0 or x 1 = 0

x = 0 2 or x = 0 + 1

x = -2 or x = 1

Documents

Unit # 2 Theory of Quadratic Equations