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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
By Quadratic Formula
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
By Quadratic Formula
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
By Quadratic Formula
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
=
Product of the roots =
( - 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
Let and be the roots of the equation
Sum of roots =
+ =
+ = -7
= -7 (i)
Product of the roots =
=
= 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
Let and be the roots of the equation
Sum of roots =
+ =
+ =
= (i)
Product of the roots =
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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
Let and be the roots of the equation
Sum of roots =
+ =
Product of the 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)
Let and be the roots of the equation
Sum of roots =
+ =
+ =
Product of the 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 = Product of roots
P = (2) (-6)
P = -12
The required equation is
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 = Product of roots
P = (-1) (-7)
P = 7
The required equation is
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 = Product of roots
P = (1+ ) (1 )
P = (1)2 ( )
2
P = 1 (-1)
P = 1 + 1
P = 2
The required equation is
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 = Product of roots
P = (3 + ) (3 - )
P = (3 )2 ( )2
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P = 9 2
P = 7
The required equation is
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
Product of the roots =
=
= 6
Given roots are and
S= +
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S =
S =
S =
S =
S =
S =
P = ( ) ( )
P = 1
The required equation is
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
The required equation is
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
The required equation is
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
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