Quadratic Equations

329

Quadratic Equations

Chapter-15

Quadratic Equations
Important Definitions and Related Concepts
1. Quadratic Equation
If p(x) is a quadratic polynomial, then p(x) = 0 is called a quadratic equation. The general formula of a quadratic equation is ax 2 + bx + c = 0; where a, b, c are real numbers and a  0. For example, x2 – 6x + 4 = 0 is a quadratic equation.

2. Roots of a Quadratic Equation
Let p(x) = 0 be a quadratic equation, then the values of x satisfying p(x) = 0 are called its roots or zeros.
For example, 25x2 – 30x + 9 = 0 is a quadratic equation.
3
And the value of x = is the solution of the given
5
equation.
3
Since, if we put x = in 25x2 – 30x + 9 = 0, we have,
5
2

3
 3
LHS = 25 ×   – 30 ×
+ 9
 5
5
= 9 – 18 + 9 = 0 = RHS
Finding the roots of a quadratic equation is known as solving the quadratic equation.

5. Methods of Solving Quadratic Equation
( i ) By Factorization
This can be understood by the examples given below: 2
Ex. 1: Solve: 25 x  30 x  9  0

Soln: 25x 2  30x  9  0 is equivalent to

5x 2  25x   3  32

0

 5 x  32  0

3 3
3
, or simply x  as the
5 5
5
required solution.
This gives x 

Ex. 2: Find the solutions of the quadratic equation

x 2  6 x  5  0 and check the solutions.
2
Soln: The quadratic polynomial x  6x  5 can be factorised as follows:

x 2  6x  5  x 2  5x  x  5
= x (x  5 )  1(x  5)

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3. Quadratic Formulae
(i)

If  and  are the two roots of ax2 + bx + c = 0, then  b  b 2  4ac

2a

 b  b 2  4ac and  
2a
b
( i i) Sum of the roots (  )   a c
(iii) Product of the roots ()  a ( iv) A quardratic equation whose roots are  and
 is given by x 2  (  )x    0 , ie x2 –(sum of roots)x + product of roots = 0
(v) In ax 2 + bx + c = 0, t he expr ession
D = b 2 – 4ac is called its discriminant.

4. Nature of Roots of ax2 + bx + c = 0
Let D = b 2 – 4ac be the discriminant of the given equat ion. Then r oots of t he equat ion ax 2 + bx
+ c = 0 are
(i)
real and equal if D = 0.
( i i) real, unequal and rational, when D > 0 and D is a perfect square.
( i ii ) real, unequal and irrational, when D > 0 and D is not a perfect square.
( iv) imaginary, if D < 0.
(v) integers, when a = 1, b and c are integers and the roots are rational.

= (x  5 )(x  1)
Ther ef or e, t he giv en quadr atic equation becomes (x  5)(x  1)  0
This gives x = – 5 or, x = – 1
Therefore, x = – 1, – 5 are the required solutions of the given equation.
Check: We substitute x = – 1 and x = – 5 in the given equation and get
(i) (–1)2 + 6(–1) + 5
= 1 – 6 + 5 = 0
(ii) (–5)2 + 6(–5) + 5
= 25 – 30 + 5 = 0
Therefore, the solutions are correct.

Ex. 3: Solve:
2x
1
3x  9


0
x  3 2 x  3 x  3 2 x  3 
Soln: Obv iously, t he given equation is valid if x  3  0 and 2x  3  0 .
Multiplying throughout by x  32x  3 , we get, 2x 2x  3   1x  3  3x  9  0
 4x 2  10x  6  0
 2x 2  5x  3  0
 2x 2  2x  3x  3  0
 2x  3x  1  0
But 2x + 3  0, so we get x + 1 = 0.
This gives x = –1 as the only solution of the given equation.

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SSC Advanced Maths

( i i) By Using Method of Completing Square
It is not always easy to factorise polynomials and solve quadratic equations as discussed above. For example, consider the quadratic equation x 2 + 5x
+ 5 = 0. If we want to factorise the left-hand side of the equation using the method of splitting the middle term, we must determine two integral factors of 5 whose sum is 5. But, the only factors of
5 are 1 and 5 or –1 and –5. In both the cases, the sum is not 5. Therefore, using factorisation, we are unable to solve the quadratic equation x2 + 5x
+ 5 = 0. Here, we shall discuss a method to solve such quadratic equations. Let us consider the following example:
Ex. 4: Solve: x2 + 3x + 1 = 0.
Soln: We have x2 + 3x + 1 = 0
We add and subtract (

1 coefficient of x) 2
2

in LHS and get
2

2

 b  b 2  4ac
2a

=

 5  (5 )2  4  2  (6)
22

=

 5  25  48
 5  73
=
4
4

 5  73
 5  73
,
4
4
Ex. 6: Determine whether the following quadratic equations have real roots and if they have, find them:

=

(b) x 2  4 x  4  0

2

 3
 3
 3
2
 x  2  x       1  0
 2
 2
 2

1
3
x
0
3
2
Soln: (a) Comparing t he giv en equat ion wit h ax2 – bx + c = 0, we find that, here a = 3, b = – 5 and c = 2.
Therefore, discriminant D = b 2 – 4ac
= (–5)2 – 4 × 3 × 2 = 1
Since D > 0, the equation has two distinct roots say, ,  given by
2
5 1
5 1

, ie, 1,  
, ie,
3
6
6
2
(c) x 

2

3
5

  x     0
2
4
2
 5
3

  x    

2
 2 

x =

(a) 3 x 2  5 x  2  0

 3
 3 x 2  3x  1        0
 2
 2
2

Ex. 5: Solve the following equation:
2x2 + 5x – 6 = 0.
Soln: Here, the given equation
2x2 + 5x – 6 = 0
 a = 2, b = 5, c = –6

2

3
5

2
2

K KUNDAN
This gives x 

or x 



 3 5



2

or x 

3  5
2

3  5 3  5
,
ar e t he
2
2 solutions of the given equation.
(iii) By Using Quadratic Formulae
If the equation is ax2 + bx + c = 0, then
Ther efor e, x  

 b  b 2  4ac x 
2a

 b  b 2  4ac or , x 
2a

and

is often referred to as quadratic

formula.
(a) When b2 – 4ac = 0, ie b2 = 4ac, then   

b
2a

b where  and  are the two roots
2a
of the above equation, ie both the roots are equal. (b) When b2 – 4ac > 0, ie b2 > 4ac, then the equation has two distinct real roots ,  given by and   



 b  b 2  4ac
2a

and  

 b  b 2  4ac
2a

 b 
 4 
 ie,  
  2. given by  
 2a 
 2 
2

2

 b  b  4ac
2a

2
.
3
(b) Here a = 1, b = – 4 and c = 4. Therefore,
D = (–4)2 – 4 × 1 × 4 = 16 – 16 = 0
Hence, the equation has a repeated root
Thus, the two roots are 1 and

1
53
3
 1
6
0
(c) D =     4  1 
=
9
9
2
 3
Therefore, the equation does not have real roots. Ex. 7: Det er mi ne v al ue( s) of p for whi ch t he quadratic equation 2x2 + px + 8 = 0 has real roots.
Soln: D = b2 – 4ac
= p2 – 4 × 2 × 8
= p2 – 64
For the equation having real roots, D  0, ie, p2 – 64  0
 p2 – 64, ie, p2 – 82.
This gives p  8 or p  –8.
Note: p2 – 82  0. gives
(p + 8) (p – 8)  0 .... (i)

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Quadratic Equations
(i) holds if

2

1


117 12K  3 10  K

(a) p  8  0 and p  8  0 ie, p  8 and p  8

2

  12K  3  117 10  K 

These give p  8 . or 2

 9  4K  1  117 10  K 

(b) p  8  0 and p  8  0

 16K 2  8K  1  130  13K

ie p  8 and p  8
These give p  8 .
Therefore, required values of p are p  8 or

p  8 .

6. Condition for Common Roots
Let a1x 2  b1x  c1  0 and a 2x 2  b2x  c 2  0

be two

quadratic equations such that a1, a1  0 and a1b2  a2b1.
Let  be the common root of these two equations.
Then, a1 2  b1  c1  0 and a 2 2  b2  c 2  0
Solving these two equations by cross-multiplication, we get

2

1

 b1c 2  b2c1 c1a 2  c 2a1 a1b2  a 2b1
 2 

b1c 2  b2c1 c a  c 2a1
 1 2 a1b2  a 2b1 and a1b2  a 2b1

 16K 2  5K  129  0
 16K2 – 48K + 43K – 129 = 0
 16K (K – 3) + 43(K – 3) = 0
 (16K + 43) (K – 3) = 0
K = 

43 or 3
16

 =

12K  3
= –3 or 4
10  K

7. Maximum or Minimum Value of a Quadratic
Expression
As we hav e already seen, equation of the type ax2 + bx + c = 0 (where, a  0) is called a quadratic equation. An expression of the type ax 2 + bx + c is called a “quadratic expression”.
The quadratic expression ax2 + bx + c takes different values as x takes different values.
As x varies from – to +, the quadratic expression ax2 + bx + c
( i ) has a minimum value whenev er a > 0. The minimum value of the quadratic expression is

K KUNDAN
Eliminating , we get

 b1c 2  b2c1   c1a 2  c 2a1 

 a1b2  a2b1   a1b2  a 2b1 

 4ac  b 2 




4a



2

 b1c 2  b2c1  a1b2  a 2b1   c1a 2  c 2a1 

2

The above is the required condition for the two quadratic equations to have a common root.
The common root is given by


c1a 2  c 2a1 b c  b2c1 or   1 2 a1b2  a 2b1 c1a 2  c 2a1

Note: ( i ) To find the common root of two equations, make the coefficient of second degree terms in two equations equal and subtract. The value of x so obt ained is the r equir ed common root.
( i i) If t he t wo equat ions hav e bot h r oot s a1 b1 c1 common, then a  b  c
2
2
2

Ex. 8: Find the value of K, so that the equations x 2  x  12  0 and Kx 2  10 x  3  0 m ay have one root common. Also find the common root. Soln: Let  be the common root of the two equations.
Hence, 2 –  – 12 = 0 and
K2 + 10 + 3 = 0
Solving the two equations,

and it occurs at x  

b
.
2a

( i i) has a miximum value whenev er a < 0. The miximum value of the quadratic expression is

 4ac  b 2 




4a



Ex. 9:
Soln.

and it occurs at x  

b
.
2a

Find the maximum or minimum value of
–5x2 + 20x + 40.
A quadratic expression of the form ax2 + bx + c, will have a minimum value when a > 0 and maximum value when a < 0. Its maximum or minimum value is given by

4ac  b 2 and it
4a

b
.
2a
Given, a = –5, b = 20 and c = 40
Since, a < 0, the expression has a maximum value. occurs at x  

 the maximum value =

4(5)(40 )  20 2
= 60
4(5)

Ex. 10: In the previous example, find the value of x for which the maximum value occurs.
Soln. The maximum vlaue of the expression occurs at x 

b
20

2
2a 2(5 )

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SSC Advanced Maths

Exercise
1.

a)
2.

1
8

b) 

14. If the roots of equation ax 2  bx  c  0 are   then what will be the value of 2 + 2?

2x  3 3x  1

is
2x  1 3x  1

The solution of

1
8

c)

8
3

d) 

8
3

Let f(x) = x² – 27x + 196. If f(x) = x, then the value of x is a)  28
b) 14
c) 7
d) 4
2/3

The solution of the equation x
 x  2  0 is
a) x = 2 or x = 4
b) x = 1 or x = –8
c) x = 8 or x = –1
d) x = 1 or x = 8

4.

Minimum value of 2x² – 8x + 7 is
a) 2
b) –2
c) 1

6.

8.

The solutions of the equation

25  x 2  x  1 are
b) x = 5 and x = 1
d) x = 4 and x = –3

Which of the following equations has real roots?
a) 3x² + 4x + 5 = 0
b) x² + x + 4 = 0
c) (x – 1) (2x – 5) = 0
d) 2x² – 3x + 4 = 0
Find the maximum or minimum v alue of the expression 2x – 3x2 + 7.

b 2  2ac a2 c)

a2 b  2ac

d)

a2 b  2ac

2

2

15. If the roots of the equation (c 2  ab )x 2  2(a 2  bc )x

(b 2  ac )  0 for a  0 are real and equal, then the value of a 3  b 3  c 3 is
a) abc
b) 3 abc

c) zero

d) None of these

and  are t he t wo r oots of t he equation

2x 2  7x  3  0 , then find the value of ( + 2) ( + 2)
a) 9
b) –9.5
c) 9.5
d) 6

17. The roots of 2x 2  6x  3  0 are
a) Real, unequal and rational
b) Real, unequal and irrational
c) Real and equal
d) Imaginary
18. If the roots of ax 2  bx  c  0 be equal, then the value of c is
a)

b
2a

b)

b
2a

c)

 b2
4a

d)

b2
4a

K KUNDAN
22
3

a) Minimum of

b) Minimum of

22
c) Maximum of
3

9.

b)

16. If 

d) 8 and 2

34 and  34

a) x = 3 and x = 4
c) x = –3 and x = 4
7.

d) –1

The roots of the equation (x + 3) (x – 3) = 25 are
a) 5 and –5
b) 3 and –3
c)

b 2  2ac a2 1/3

3.

5.

a)

47
8

47
d) Maximum of
8

If a and b are the roots of the equation x² – 6x + 6 = 0, then the value of a² + b² is
a) 36
b) 24
c) 12
d) 6 x  4  x  10 5
 . The value of x is x  4  x  10 2

10. Given

a) 1

b)

331
5

c)

263
20

d)

17
21

11. The roots of the quadratic equation 2x 2  11x  15  0 are 5
3
5
a) 3,
b) 5,
c) 3, 
d) None of these
2
2
2
12. If  are the roots of x 2  3x  2  0, then the equation with roots (+ 1) ( + 1) is
a) x 2  5x  6  0

b) x 2  5x  6  0

c) x 2  5x  6  0

d) x 2  5x  6  0

d) 1,

b)

3
2

3
2

e) None of these

x 2  2x  p  0 is 10, then the value of p will be
a) –3
b) 3
c) 6
d) –6

2 0 . If the equation 4x 2  x  p  1  1  0 has exactly two equal roots, then one of the values of p is
a) 5
b) –3
c) 0
d) 3

21. If  and  are the roots of the equation 5 x 2  x  2  0 ,

2
2
and



then the equation for which the roots are is a) x 2  x  10  0

b) x 2  x  10  0

c) x 2  x  10  0

d) x 2  x  10  0

2 2 . If ,  are the roots of the equation x² – q(1 + x) – r = 0, then 1   1   is
a) 1 – r

b) q – r

c) 1 + r

d) q + r

23. For what value of p, the difference of the roots of the equation x² – px + 8 = 0 is 2?
a)  2
b)  4
c)  6
d)  8
24. Find the minimum value of the quadratic equation
5x² + 7x + 2.

13. If 2a 2  a  2  1 and a > 0, then a = ?
a) 1

19. If the sum of the squares of the roots of the equation

c) 3

a) 

9
10

b) 

9
20

c) 

29
10

d) 

29
20

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Quadratic Equations
25. The roots of the quadratic equation ax² + bx + c = 0 will be reciprocal to each other if
1
a) a =
b) a = c
c) b = ac
d) a = b c 1
1
1
1
 

26. If the equation has two roots x a x k a k which are equal in magnitude but opposite in sign, then ‘a’ is equal to
a) k

b) –k

c)  2k

d) 2k

27. If the roots of the quadratic equation x² – px + q = 0 differ by unity, then
a) p² = 4q + 1
b) p² = 4q – 1
c) q² = 4p + 1
d) q² = 4p – 1
28. If  and  are the roots of the equation ax² + bx + c = 0,
1
1 then the equation whose roots are   and   is 

a) abx² + b (c + a) x + (c + a)² = 0
b) (c + a) x² + b (c + a) x + ac = 0
c) cax² + b (c + a) x + (c + a)² = 0
d) cax² + b (c + a) x + c (c + a)² = 0
29. If a³ = b³ and a  b , then the sum of the roots of the equation x² – (a² + ab + b²) x + k = 0 is equal to
a) k
b) b²
c) 0
d) a²
30. If 0 < a < 4, then the equation ax(1 – x) = 1 has
a) two equal roots
b) one positive root and one negative root
c) two irrational roots
d) No real roots

37. If one root of the quadratic equation x² + bx + c = 0 is square of the other, then b³ + c² + c equals
a) bc
b) 2bc
c) 3bc
d) 6bc
38. If t he equat ion

3x 2  7x  30  2x 2  7x  5  x  5

has x1, x 2 as its roots, then the value of x1 x 2 is
a) 15

b) 0

c) –5

d) –15

39. Which of the following expressions cannot be equal to zero, when x 2  2x  3 ?
a) x 2  7x  6

b) x 2  9

c) x 2  4x  3

d) x 2  6x  9

40. If 2x 2  7xy  3y 2  0 , then the value of x : y is
a) 3 : 2
b) 2 : 3
c) 3 : 1 and 1 : 2
d) 5 : 6
41. If ,  are t he r oot s of t he quadr atic equation

x 2  px  q  0, then 3  3 = ?
a) q 3  3 pq

b) q 3  3 pq

c) p 3  3 pq

d) p 3  3 pq

42. Find the maximum value of the expression –2x² – 8x + 5.
a) 13
b) 26
c) –6
d) –3
43. A and B solved a quadratic equation. In solving it, A made a mistake in the constant term and obtained the roots as 5, –3, while B made a mistake in the coefficient of x and obtained the roots as 1, –3. The correct roots of the equation are
a) 1, 3
b) –1 , 3
c) –1 , –3 d) 1, –1

K KUNDAN

31. The condition that one root of the equation ax² + bx + c = 0 is three times the other is
a) b² = 8ac
b) 3b² = 16ac
c) 3b² = 5ac
d) b² + 3ac = 0

32. If the equation (m – n)x² + (n – l)x + l – m = 0 has equal roots, then l, m and n satisfy
a) 2l = m + n
b) 2m = n + l
c) m = n + l
d) l = m + n

33. If one root of the quadratic equation 3x² – 10x + p = 0

1
, t hen t he v alue of p and t he ot her root
3
respectively are is a) 3,

1
3

b) 3, 3

c) –

1
1
, –
3
3

d) –3, –3

34. If the quadratic equation 2x² + 3x + p = 0 has equal roots, then the value of p will be

4
a)
3

5
b)
4

6
c)
5

9
d)
8

35. If the roots of the equation 12x² + mx + 5 = 0 are in the ratio 3 : 2, then the value(s) of m is/are
a) 6 5

b) 6 5

c) 5 10

d) 6 10

36. The set of values of k, for which x² + 5kx + 16 = 0 has no real root, is

8
a) k 
5
c) 

8
8
k 
5
5

8
b) k 
5
d) –8 < k < 8

44. If (x – 3) (2x + 1) = 0, then possible values of 2x + 1 are
a) 0 only
b) 0 and 3
1
c)  and 3
d) 0 and 7
2

15
, then x is equal to x a) –5 or –3
b) –5 or 3
c) –3 or 5
d) 3 or 5

45. If x + 8 =

10
.
a
b) 5, –2

46. Find a, if a – 3 =
a)

7, 7

c) –5, 2

d)  7, 7

47. Which of the following is a quadratic equation?
a) x2 + 2x + 1 = (4 – x)2 + 3

2

b) –2x2 = (5 – x)  2x  
5

3
c) (k + 1)x2 + x = 7, where k = –1
2
d) x3 – x2 = (x – 1)3
48. Which of the following equations has 2 as a root?
a) x2 – 4x + 5 = 0
b) x2 + 3x – 12 = 0
2
c) 2x – 7x + 6 = 0
d) 3x2 – 6x – 2 = 0

1 is a root of the equation x2 + kx –
2
the value of k is
1
a) 2
b) –2
c)
d)
4

49. If

5
= 0, then
4
1
2

334

SSC Advanced Maths

50. Value(s) of k for which the quadratic equation 2x2 – kx
+ k = 0 has equal roots is/are
a) 0
b) 4
c) 8
d) 0, 8
51. If x is real, then the minimum value of (x2 – x + 1) is
3
1
a)
b) 0
c) 1
d)
4
4
52. Which constant must be added and subtracted to solve
3
the quadratic equation 9x 2 + x – 2 = 0 by the
4
method of completing the square?
1
1
1
9
a)
b)
c)
d)
8
64
64
4
53. The quadratic equation 2x2 –
a) two distinct real roots
b) two equal real roots
c) no real roots
d) More than 2 real roots

5 x + 1 = 0 has

the roots of x2 – 3kx + 2k2 – 1 = 0 is
What is the nature of roots? positive negative not integral

65. What are the roots of the equation (a + b + x)–1 = a-1 + b–1 + x–1?
a) a, b
b) –a, b
c) a, –b
d) –a, –b
66. What is one of the values of x in the equation

a)

d) 5x2 – 3x + 1 = 0

55. Which of the following equations has no real roots?
a) x2 – 4x + 3 2 = 0

b) x2 + 4x – 3 2 = 0

c) x2 – 4x – 3 2 = 0

d) 3x2 +4 3 x + 4 = 0

56. (x2 + l)2 – x2 = 0 has
a) four real roots
c) no real roots

64. The product of
7 for a fixed k.
a) Integral and
b) Integral and
c) Irrational
d) Rational but

x
1  x 13


?
1 x x 6

54. Which of the following equations has two distinct real roots?
9
a) 2x2 – 3 2 x +
= 0
b) x2 + x – 5 = 0
4
c) x2 + 3x + 2 2 = 0

63. For what value of k, does the equation [kx2 + (2k + 6)x
+ 16 = 0] have equal roots?
a) 1 and 9
b) –9 and 1
c) –1 and 9
d) –1 and –9

b) two real roots
d) One real root

5
13

b)

7
13

c)

9
13

d)

11
3

67. If the equations 2x2 – 7x + 3 = 0 and 4x2 + ax – 3 = 0 have a common root, then what is the value of a?
a) –11 or 4
b) –11 or –4
c) 11 or –4
d) 11 or 4
68. If one root of px2 + qx + root, then which one of
a) 2q2 = 9pr
c) 4q2 = 9r
69.

r = 0 is double of the other the following is correct?
b) 2q2 = 9p
d) 9q2 = 2pr

If 3x

+ 27(3–x) = 12, then what is the value of x?
a) 1 only b) 2 only
c) 1 or 2
d) 0 or 1

70. What is the magnitude of difference of the roots of x2 – ax + b = 0?

K KUNDAN

57. If x 2 – kx – 21 = 0 and x2 – 3kx + 35 = 0 have one common root, then what is the value of k?
a) +4 only
b) –4 only
c) ± 4
d) ±1
58. How many r eal v alues of x sat isfy t he equat ion
2
x3

1 x3 
20 ?
a) Only 1 value
c) 3 values

b) 2 values
d) No value

59. If sum of the roots of the equation ax2 + bx + c = 0 is equal to the sum of their squares, then which one of the following is correct?
a) a2 + b2 = c2
b) a2 + b2 = a + b
c) 2ac = ab + b2
d) 2c + b = 0
60. If ,  are the roots of x2 – 5x + k = 0, then what is the value of k such that  –  = 1?
a) 1
b) 3
c) 4
d) 6
61. Which one of the following is the equation whose roots are respectively three times the roots of the equation ax + bx + c = 0?
a) ax2 + 3bx + c = 0
b) ax2 + 3bx + 9c = 0
c) ax2 – 3bx + 9c = 0
d) ax2 + 3bx + 3c = 0
62. If ,  are the roots of the equation (ax + bx + c = 0), then what is the value of 3 + 3?
a)

c)

b 3  3abc a 3

3abc  b 3 a3 b)

d)

a3  b3
3ab

b 3  3abc a 3

a)

a 2  4b

c) 2 a 2  4b

b)

b 2  4a

d)

b 2  4ab

71. Which one of the following is the quadratic equation whose roots are reciprocal to the roots of the quadratic equation 2x2 – 3x – 4 = 0?
a) 3x2 – 2x – 4 = 0
b) 4x2 + 3x – 2 = 0
c) 3x2 – 4x – 2 = 0
d) 4x2 – 2x – 3 = 0

72. The v alue of y which will sat isf y the equations
2x2 + 6x + 5y + 1 = 0 and 2x + y + 3 = 0 may be found by solving which one of the following equations?
a) y2 + 14y – 7 = 0
b) y2 + 8y + 1 = 0
c) y2 + 10y – 7 = 0
d) y2 – 8y + 7 = 0
73. If a polynomial equation has rational coefficients and has exactly three real roots, then what is the degree of the polynomial?
a) Equal to 3
b) Greater than or equal to 3
c) Strictly greater than 3
d) Less than 3
74. If ,  are the roots of ax2 + bx + c = 0, then what is

 1
1  the value of  2  2  ?




2

a)

b (b 2  4ac ) c 4

b)

2

c)

(b  4ac ) c2 b(b 2  4ac ) c2 2

d)

(b  4ac ) c4 335

Quadratic Equations
75.

What

is

2x

3x

one

of

t he

r oot s

of

t he

equation

87. W hat

3x
3
 ?
2x
2

a) 1

b) 2

is

x x3 c) 3

d) 4

76. If ,  are the roots of the equation (x2 – 3x + 2 = 0), then which equation has the roots ( + 1) and ( + 1)?
a) x2 + 5x + 6 = 0
b) x2 – 5x – 6 = 0
c) x2 + 5x – 6 = 0
d) x2 – 5x + 6 = 0

t he

solut ion

of

t he

equat ion

x 3
3


? x 2

a) 1
c) 4

b) 2
d) None of these

88. What are the roots of the equation 4x – 3.2x+2 + 32 = 0?
a) 1, 2
b) 3, 4
c) 2, 3
d) 1, 3

77. If one of the roots of the equation ax2 + x – 3 = 0 is
–1.5, then what is the value of a?
a) 4
b) 3
c) 2
d) –2

89. If  and  are the roots of the equation x2 – x – 1 = 0, then what is the value of (4 + 4)?
a) 7
b) 0
c) 2
d) None of these

78. When the roots of the quadratic equation ax2 + bx + c
= 0 are negative of reciprocals of each other, then which one of the following is correct?
a) b = 0
b) c = 0
c) a = c
d) a = –c

90. If sum as well as product of roots of a quadratic equation is 9, then what is the equation?
a) x2 + 9x – 18 = 0
b) x2 – 18x + 9 = 0
c) x2 + 9x + 9 = 0
d) x2 – 9x + 9 = 0

79. W hat ar e the root s of t he equat ion ( x 2 – 6x
+ 45) = 100
a) 9, –5
b) –9, 5
c) 11, –5
d) –11, 5

91. What is the least integral value of k for which the equation x2 – 2(k – 1)x + (2k + 1) = 0 has both the roots positive? 1
1
1

 x a x b c is zero. What is the product of the roots of the equation?

1
c) 4
d) 0
2
92. If  and  are the roots of the equation x2 – 6x + 6 = 0, what is 3 + 3 + 2 + 2 +  +  equal to?
a) 150
b) 138
c) 128
d) 124

80. The sum of the roots of the equation

a) 

(a  b )
2
2

c) 

b)
2

(a  b )
2
2

(a  b )
2

d)

b) 

a) 1

93. W hat ar e the root s of t he quadr at ic equat ion a2b2x2 – (a2 + b2)x + 1 = 0 ?

2

(a  b )
2

a)

81. For what value of k, will the roots of the equation kx2 – 5x + 6 = 0 be in the ratio of 2 : 3?
a) 0
b) 1
c) –1
d) 2

1 a2 ,

1

b) 

b2

1 a2 , 

1 b2 K KUNDAN

82. What is the ratio of sum of squares of roots to the product of the roots of the equation 7x2 + 12x + 18 = 0?
a) 6 : 1
b) 1 : 6
c) –6 : 1
d) –6 : 7 x (x  1)  (m  1) x

(x  1)(m  1) m equal, then what is the value of m?

83. If the roots of the equation

a) 1

b)

1
2

c) 0

d) 

are

1
2

84. If  and  are the roots of the equation x2 + px + q = 0, then – –1 , ––1 ar e t he roots of which one of t he following equations?
a) qx2 – px + 1 = 0
b) q2 + px + 1 = 0
c) x2 + px – q = 0
d) x2 – px + q = 0
85. If one root of the equation ax2 + x – 3 = 0 is –1, then what is the other root?
1
1
3
a)
b)
c)
d) 1
4
2
4
86. If the equation (a2 + b2)x2 – 2 (ac + bd) x + (c2 + d2) = 0 has equal roots, then which one of the following is correct? a) ab = cd
b) ad = bc
c) a2 + c2 = b2 + d2
d) ac = bd

c)

1

a2

, 

1

b2

d) 

1

a2

,

1

b2

94. If one root of the equation 2x2 + 3x + c = 0 is 0.5, then what is the value of c?
a) –1
b) –2
c) –3
d) –4

95. What is the condition that the equation ax2 + bx + c =
0, where a  0 has both the roots positive?
a) a, b and c are of same sign
b) a and b are of same sign
c) b and c have the same sign opposite to that of a
d) a and c have the same sign opposite to that of b

96. The equation (1 + n2)x2 + 2ncx + (c2 – a2) = 0 will have equal roots, if
a) c2 = 1 + a2
b) c2 = 1 – a2
2
2
2
c) c = 1 + n + a
d) c2 = (1 + n2)a2
97. The equation whose roots are twice the roots of the equation x2 – 2x + 4 = 0 is
a) x2 – 2x + 4 = 0
b) x2 – 2x + 16 = 0
2
c) x – 4x + 8 = 0
d) x2 – 4x + 16 = 0
3
98. If x2 – 4x +1 = 0, then what is the value of x 

a) 44

b)48

c) 52

d) 64

1 x3 ?

336

SSC Advanced Maths

Answers and explanations
1. b;

2x  3 3x  1

2x  1 3x  1
(2x + 3) (3x + 1) = (2x – 1) (3x – 1)
 6x² + 11x + 3 = 6x² – 5x + 1

8. c;

The given equation is

1
8
2. b; f(x) = x² – 27x + 196 and f(x) = x means x² – 27x + 196 = x
 x² – 28x + 196 = 0
 (x – 14)² = 0  x = 14.

 The expression has maximum value =

 16x  2  x  

3. b;

1

Let x 3 be y.
2
3

1
3

Then, x  x  2  0
 y² + y – 2 = 0
 y² + 2y – y – 2 = 0
 y(y + 2) –1(y + 2) = 0
 (y – 1)(y + 2) = 0 y = 1 or –2.
Therefore, x = y³ gives x = 1 or –8
4. d; 2x² – 8x + 7
= 2 (x² – 4x) + 7
= 2 (x² – 4x + 4 – 4) + 7
= 2[(x – 2)² – 4] + 7
= 2 (x – 2)² – 8 + 7 = 2 (x – 2)² – 1
Now 2 (x – 2)² is positive quantity.
 For minimum value of 2x² – 8x + 7 = 2 (x – 2)² – 1,
2 (x – 2)² = 0 x = 2
 The minimum value of 2x² – 8x + 7 is –1.
Alternative Method:
The given expression = 2x² – 8x + 7
Here, a = 2 > 0
 The given expr ession has minimum v alue

The given expression = –3x2 + 2x + 7
Here, a = –3 < 0

4  (3)  7  (2)2 22

4  (3)
3

 Maximum value =
9. b;

4ac  b 2
4a

We know that sum of the roots = 

b a and the

c product of the roots =
.
a
6
 6 
 = 6 and ab =
a + b =  
= 6
1
1


a² + b² = (a + b)² – 2ab = 36 – 12 = 24
10. c; The given expression


x  4  x  10 5

x  4  x  10 2



x  4  x  10

x  4  x  10







x  4  x  10



 x  4   x  10

x  4  x  10 5

x  4  x  10 2

2



5
2

x  4  x  10  2 x 2  6x  40 5

14
2

K KUNDAN
=

5. c;

56  64
8
4ac  b 2
4  2  7  (8)2
=
=
=
= –1
8
8
4a
42

x² – 9 = 25 x² = 34
 x   34

6. d;

7. c;

 2x  6  2 x 2  6x  40  35
 2x  2 x 2  6x  40  41

2
2
  2x  41   2 x  6x  40 

4x² + 1681 – 164x = 4(x² – 6x – 40)
4x² + 1681 – 164x = 4x² – 24x – 160
263
 140x = 1841  x 
20

11. a; The given quadratic equation = 2x 2  11x  15  0

2

25  x  x  1
On squaring both sides, we get,
25 – x² = (x – 1)²
25 – x² = x² + 1 – 2x
2x² – 2x – 24 = 0
x² – x – 12 = 0
x² – 4x + 3x – 12 = 0
x(x – 4) + 3(x – 4) = 0
(x – 4) (x + 3) = 0 x = 4, x = –3
The equation ax² + bx + c = 0 has real roots if b2 – 4ac > 0.
Option (a)  b² – 4ac
= (4)2 – 4 × 3 × 5
= 16 – 60 = –44 < 0
Option (b) : b² – 4ac
= (1)2 – 4 × 1 × 4
= 1 – 16 = –15 < 0
Option (c) : (x – 1) (2x – 5) = 0
2x² – 7x + 5 = 0
b² – 4ac = 49 – 40 = 9 > 0
Option (d) : b² – 4ac
= (–3)2 – 4 × 2 × 4 = 9 – 32 = –23 < 0

2

x =

11  1
5
11  121  120
 3,
=
4
2
4

12. d;  +  = 3 and  = 2
Sum of the roots of the required equation
= ( + 1) + ( + 1)
=  +  + 2 = 5
Product of the roots = ( + 1)( + 1)
=  +  +  + 1
= 2 + 3 + 1 = 6
 The equation, whose roots are ( + 1) and ( + 1), is given by x2 – 5x + 6 = 0
  Quadratic equation = x 2 – (Sum of roots)x + product of roots = 0]
13. a; 2a2 + a – 2 = 1
 2a2 + a – 3 = 0
 2a2 + 3a – 2a – 3 = 0
 a (2a + 3) –1 (2a + 3) = 0
 (2a + 3)(a – 1) = 0
3
 a = 1, a = 
2
Since a > 0, therefore a = 1 is the only possibility.

337

Quadratic Equations

22. a; Given that  and  are the roots of the equation x² – qx – (q + r) = 0

b c and   . a a b 2 2c b 2  2ac

 2  2  (  )2  2 = 2  a a a2 14. a; We know that     

     q and   (q  r )

2
2
15. b; Discriminant = [ 2(a 2  bc ) ]2 –4 (c  ab )(b  ac ) = 0

for the roots to be equal.
3
3
2
 a 4  b 2c 2  2a 2bc  c 2b 2 + ac  ab  a bc  0

23. c; The given equation is x² – px + 8 = 0.
If the roots are  and , then     p and   8
2

2

            4  p 2  32

 a 3  2abc  c 3  b 3  abc  0

 2² = p² – 32
 p² = 36 p =  6

 a 3  b 3  c 3  3abc
16. c;    

Now, (1   )(1  )  1      
= 1 + q – (q + r ) = 1 – r

24. b; The given expression = 5x 2 + 7x + 2
Here, a = 5 > 0

7
3
and  
2
2

 (  2)(  2) =   2  2  4

 The expression has minimum value =

=   2(  )  4
=

 Minimum value =

3
19
 11 
 9.5
2
2
2

17. b; Discriminant (D) = b – 4ac = 36 – 24 = 12 > 0
 Roots are real and distinct but irrational.
18. d; b2 – 4ac = 0  c =

b2
4a

4  5  2  (7)2
9

45
20

1 be the roots of the given equation.


25. b; Let  and
  

4ac  b 2
4a

1 c

 c = a.
 a

26. c; Clearly x = k is a root of the equation

19. b; Let the roots of the equation x² + 2x – p = 0 be and .
Then,  +  = –2,   = –p and  2 + 2 = 10
2

       2  2  2

(–2)² = 10 + 2 × (–p)  p = 3

1
1
1
1
 

x a x k a k
Therefore, another root would be x = –k.
Putting x = –k, we have,
1
1
1
1
 

k  a k k  a k

K KUNDAN

20. d; Let the roots of the equation 4x² + x (p + 1) + 1
= 0 be  and  Then the sum of the roots

  

p 1 p 1
   
8
4

Product of the roots     

....(i)

1
4

...(ii)

From (i) and (ii), we have,
2

1
 p  1
 
 
8 
4
 (p + 1)² = 16
 p + 1 =  4  p = 3, –5
21. d; Let and  be the roots of the equation 5x² – x – 2
= 0
1
2
 + = and = 
5
5

   
2 2
Now, sum of the roots =     2   
 1 
 5 
= 2  2   1
 
 5

2
2
4
4



  10
2




5
 Required equation will be x² – (Sum of the roots)x + Product of the roots = 0
 x² + x – 10 = 0
Product of the roots 

 a 2  2k 2  a   2k

27. a; Let  and  be the roots of the equation x² – px + q = 0
Given,  –  = 1



  2  4  1

Now,  + = Sum of the roots = p and
 = Product of the roots = q
 p² – 4q = 1
 p² = 4q + 1

b c and   a a
Sum of the roots of the required equation

28. c;     


1 
1

            

 




b b b c  a 
  a c ac 1

1 
1
2
Product of the roots          =  


 



c a
a  c 
 2 a c ac 2

 Required equation is
2

a  c   0
 b(c  a ) x2  
 x  ac
 ac cax² + b(c + a) x + (a + c)² = 0

338

SSC Advanced Maths

29. c; Given, a³ = b³
a³ – b³ = 0
(a – b) (a² + ab + b²) = 0

37. c; Let  and  2 be the roots of the given quadratic equation. Then,    2  b and  3  c

 a  b  a 2  ab  b 2  0

Now,

a 2  ab  b 2
0
 Sum of the roots 
1

 c  c  3c  b    b 
38. d; Let A stand for

 a (a  4)  0

 a  0 or a  4
 If 0 < a < 4, then the given equation has no real roots. 31. b; If one root is  , then the other root is 3 .





3

 c  c 2  b 3  3bc

For real roots, a 2  4a  0

b a 

  3   6  3. 2    2

2

30. d; The given equation = ax(1 – x) = 1
 ax – ax² = 1
ax² – ax + 1 = 0

   3  

2 3

   

b
4a

....(i)

c c   3   3 2  a a
Eliminating  from (i) and (ii), we get,

....(ii)

3x 2  7x  30 and B stand for

2x 2  7x  5 .
Then we are given that
A – B = x – 5
....(i)
A2 – B2 = (3x² – 7x – 30) – (2x² – 7x – 5) = x² – 25
(A – B)(A + B) = (x – 5)(x + 5)
(x – 5)(A + B) = (x – 5)(x + 5)
A + B = x + 5
....(ii)
Adding equations (i) and (ii), we get, A = x
 2x² – 7x – 30 = 0
If x1 and x 2 are the roots, then x1x 2 

30
 15
2

39. a; x 2  2x  3  (x  3)(x  1) = 0
 (x – 3) = 0 or (x + 1) = 0

2

c
 b 
3
  3b 2  16ac
 4a  a Option (a) : x 2  7 x  6  (x  6)(x  1)

32. b; Since, (m – n)x² + (n – l)x + l – m = 0 has equal roots.
(n – l)² = 4(m – n) (l – m)
n² + l² – 2nl = 4ml – 4nl – 4m2 + 4nm
n² + l² – 2nl + 4nl – 4ml + 4m2 – 4nm
n² + l² + 2nl – 4m (l + n) + 4m2
(l + n)² – 4m(l + n) + 4m² = 0
[(l + n) – 2m]² = 0 2m = l + n

Option (b) : x 2  9  (x  3)(x  3)
Option (c) : x 2  4x  3  (x  3)(x  1)
Option (d) : x 2  6x  9  (x  3)2

K KUNDAN

33. a; Given equation is 3x² – 10x + p = 0.

40. c; 2

x2 x 7 30 y2 y

2

 1
 1
 3    10    p  0
 3
 3



 p 3

 x : y = 3 : 1 and 1 : 2

 Given equation becomes 3x² – 10x + 3 = 0.
3
 Product of the roots   1
3
 If 3 is one root, then the other root is

41. d;     p and   q

  3  3  (  )(2  2   )
= (   )[(  )2  3 )]

1
.
3

34. d; For equal roots, b² – 4ac = 0

 9  8p  0  p 

x 7  49  24
75
1


 3, y 22
4
2

= p( p 2  3q )  p 3  3 pq
42. a; The given expression = –2x2 – 8x + 5
Here, a = –2 < 0

9
8

 The expression has maximum value =

35. c; Let the roots be 3 and 2 .
2

 5 

m
5
5
 m 
, 6 ² 
 6 
 
12
12
60 
12

 m 2  250  m  5 10

36. c; For no real solution, the discriminant should be negative. (5k)2 – 4 × 1 × 16 < 0
25k2 – 64 < 0
64
8
8
 k2 <
 k < or k > –
25
5
5
8
8
 –
< k <
5
5

Maximum value =

4  (2)  5  (8)2
4  ( 2)

4ac  b 2
4a

= 13

43. b; A’s quadratic equation = x2 – (5 – 3)x + (5) × (–3)
= x2 – 2x – 15 = 0
B’s quadratic equation = x2 – (1 – 3)x + (1) × (–3)
= x2 + 2x – 3 = 0
Since A made a mistake in the constant term and
B made a mistake in coefficient of x, therefore, the correct equation is x2 – 2x – 3 = 0

 (2)  ( 2)2  4(1)(3)

24
2  4  12
=
=
2
2 1
2
= 3 and –1
 Roots of the correct equation are 3, –1.
Roots =

339

Quadratic Equations
44. d; (x – 3) (2x + 1) = 0
 x – 3 = 0 or 2x + 1 = 0.
If x = 3, then 2x + 1 = 2 × 3 + 1 = 7
45. a; The given expression is x + 8 = 

15 x  x 2  8x  15  0
x2 + 5x + 3x + 15 = 0
x + (x + 5) + 3(x + 5) = 0
(x + 5) (x + 3) = 0
 x = –5 or –3.
46. b; a 2  3a  10  0

3  9  40 3  7

2
2
 a = 5, and –2.
 a

47. d; Option (a) : x2 + 2x + 1 = (4 – x) + 3
 x2 + 2x + 1 = 16 + x2 – 8x + 3
 10x – 18 = 0 which is not of the form ax2 + bx + c, a  0. Thus, the equation is not quadratic.
Option (b) :

2

–2x2 = (5 – x)  2x  5 



2x
 –2x2 = 10x – 2x2 – 2 +
5
 50x + 2x – 10 = 0
 52x – 10 = 0 which is also not a quadratic equation.
Option (c) :
3
x2 (k + 1) + x = 7
2
Given, k = –1
3
 x2 (–1 + 1) + x = 7
2
 3x – 14 = 0 which is also not a quadratic equation.
Option (d) : x3 – x2 = (x – 1)3
 x3 – x2 = x – 3x2 (1) + 3x (1)2 – (1)3
[  (a – b)3 = a3 – b3 + 3ab2 – 3a2b]
x3 – x2 = x3 – 3x2 + 3x – 1
 –x2 + 3x2 – 3x + 1 = 0
 2x2 – 3x + 1 = 0 which represents a quadratic equation because it has the quadratic form ax2 + bx + c = 0, a  0.

1 k 5
1  2k  5
+
–
= 0 
= 0
4
4
2
4
 2k – 4 = 0  2k = 4  k = 2


50. d; Given equation is 2x2 – kx + k = 0
On comparing with ax2 + bx + c = 0, we get, a = 2, b = –k and c = k
For equal roots, the discriminant must be zero. ie, D = b2 – 4ac = 0
(–k)2 – 4(2)k = 0
 k2 – 8k = 0
k(k – 8) = 0
k = 0, 8
51. a; For expression ax 2 + bx + c, a > 0, the minimum
4ac  b 2
4a
Here, for x2 – x + 1; a = 1, b = –1, c = 1

value =

4 1 1  1 3

4 1 1
4

 Minimum value =

52. b; Given equation is 9x2 +

(3x)2 +

1
(3x) –
4

 (3x)2 + 2(3x) ×
2

3 x –
4

2 = 0

2 = 0

1
1
+  
8
8

2

2

1
–   –
8

2 = 0

K KUNDAN
48. c; We know that x =  is the root of the equation ax2 + bx + c = 0. then x =  must satisfy this equation. ie, a2 + b + c = 0
Similarly, from option (c), if x = 2 is the root of the equation 2x2 – 7x + 6 = 0.
Then, 2(2)2 – 7(2) + 6 = 2(4) – 14 + 6
= 8 – 14 + 6 = 14 – 14 = 0
x = 2 is root of the equation 2x2 – 7x + 6 = 0.

1 is a root of the quadratic equation
2
5 x2 + kx –
= 0
4

49. a; Since,

1
  
2

2

1
5
+ k  –
= 0
4
 2

1

  3x  
8


1

  3x  
8


–

1
–
64

=

64 2  1
64

2

1
Clearly,  
8

2

=

2 = 0

1 must be added and subtracted
64

to get the required answer.
53. c; Given equation is 2x2 –

5 x + 1 = 0.

On comparing with ax2 + bx + c = 0, we get, a = 2, b = – 5 and c = 1
 Discriminant, D = b2 – 4ac
= (– 5 )2 – 4 × (2) × (1)
= 5 – 8 = –3 < 0
Since, discriminant is negative, therefore, quadratic equation 2x 2 –

5 x + 1 = 0 has no real roots, ie

imaginary roots.
54. b; Let the given equation be x2 + x – 5 = 0 from option (b).
On comparing with ax2 + bx + c = 0, we get, a = 1, b = 1 and c = –5
The discriminant of x2 + x – 5 = 0 is D = b2 – 4ac
= (1)2 – 4 × (1) × (–5)
= 1 + 20 = 21
 b2 – 4ac > 0
 x2 + x – 5 = 0 has two distinct real roots.

340

SSC Advanced Maths

55. a; Let the given equation is x 2 – 4x + 3 2 = 0 from

Hence, t wo r eal v alues of x satisf y t he giv en equation. option (a).
Now, on comparing with ax2 + bx + c = 0, we get,

59. c; Let ,  be the roots of the equation ax2 + bx + c = 0.

a = 1, b = –4 and c = 3 2

 Sum of roots = ( + ) = 

The discriminant of x2 – 4x + 3 2 = 0 is D = b2 – 4ac
= (–4)2 – 4 × (1) × (3 2 )
= 16 – 12 2 = 16 – 12 × 1.41
= 16 – 16.92 = –0.92
 b2 – 4ac < 0

c a By the given condition,
 +  = 2 + 2
  +  = ( + )2 – 2 roots = () =

 x2 – 4x + 3 2 = 0 has no real roots.

2

 

56. c; Given equation is
(x2 + 1)2 – x2 = 0
 x4 + 1 + 2x2 – x2 = 0
[  (a + b)2 = a2 + b2 + 2ab]
 x4 + x2 + 1 = 0
Let, x2 = y
(x2)2 + x2 + 1 = 0
y2 + y + 1 = 0
Now, on comparing with ay2 + by + c = 0, we get, a = 1, b = 1 and c = 1
Discriminant, D = b2 – 4ac
= (1)2 – 4 × (1) × (1)
= 1 – 4 = –3
Since,
D < 0
y2 + y + 1= 0, ie x4 + x2 + 1 = 0 or (x2 + 12) – x2 = 0 has no real roots.
57. c; Let the common root of both the equations be , then 2 – k– 21 = 0
...(i)
and 2 – 3k + 35 = 0
...(ii)
Solving equations (i) and (ii), we get,

b
 and product of a b
 b
c 
     2  a  a
a 

 –ba = b2 – 2ca
 2ac = b2 + ab
60. d; Since, and  are the roots of x2 – 5x + k = 0
Then,  +  = 5,   = k
Given,  –  = 1
Then, ( – ) 2 = 1
 2 – 2 – 2 = 1
( + )2 – 4 = 1
On putting the values of  +  and , we get,
 52 – 4(k) = 1
 25 – 4k = 1 –4k = –24  k = 6
61. b; Let  and  be the roots of the equation ax2 + bx + c
= 0

c b and  = a a
Now, let the roots of the required equation be
3 and 3.
Sum of roots = 3 + 3 = 3( + )
Then,  +  = –

K KUNDAN

1
2
=
=
 21  35
 3k  k
 35k  63k

1
2

=
=
 56k
 2k
 98k


1
2
=
 2k
 98k

 2 = 49 and  =

 28 

Then, 
 k 

 b 
3b
= 3 a  = –

 a and product of roots = 3 × 3 = 9 

9c a Thus, required equation is x2 – (Sum of roots)x + (Product of roots) = 0
=

56
28
=
 2k k 2

= 49

3b
9c
x
0
a a ax2 + 3bx + 9c = 0
 x2 

62. c; Since,  and  are the roots of the equation ax2 + bx + c = 0, then

28  28
= k2
49
 16 = k2  k = ±4


b c and  = a a
3+ 3 = ( + )3 – 3( + )
+ = 

2

1

58. b; Given equation is x 3  x 3  2  0
2

1
 1
 x3  x3  2  0
 

1

 x2 + x – 2 = 0, where  x  x 3 





 It is a quadratic equation in x.
 Discriminant of x2 + x – 2 = 0 is b2 – 4ac = 12 – 4(1)(–2) = 9 > 0

 b
=  
 a
= 

b3 a 3

3

+

c  b
– 3 a   c 
 

3bc a 2

=

3abc  b 3 a3 63. a; Given equation is kx2 + (2k + 6)x + 16 = 0
Given equation has equal roots, if b 2 – 4ac = 0
 (2k + 6)2 – 4k × 16 = 0
 4k2 + 24k + 36 – 64k = 0

341

Quadratic Equations
 4k2 – 40k + 36 = 0
 k2 – 10k + 9 = 0
 k2 – 9k – k + 9 = 0
k (k – 9) – 1(k – 9) = 0
 (k – 1)(k – 9) = 0
 k = 1 and 9

2

64. c; Let  and  be the roots of the equation x2 – 3kx + 2k2 – 1 = 0
  = 2k2 – 1
But,  = 7
 2k2 – 1 = 7
 2k2 = 8 k2 = 4 k = ± 2
On putting k = ±2 in the given equation, we get, x2 ± 6x + 7 = 0

b 2  4ac =

Now,

62  4  7 =

36  28 = 2 2

Hence, roots of given equation are irrational.
65. d; Given,



1
1 1 1
   a b  x a b x

1
1 1 1
   a b  x x a b

(a  b )
(a  b )

=
(a  b  x )x ab x2 + (a + b)x + ab = 0
 (x + a)(x + b) = 0
 x = –a, –b
66. c; Let

x
y
1 x

1
1
4   a    3  0
2
2 a a
 0
 1  3  0 
2
2
a = 4
When x = 3
4(3)2 + a(3) – 3 = 0
 36 + 3a – 3 = 0 a = –11
a = –11 or 4

68. a; Given, px2 + qx + r = 0
Let the roots be  and .
By the given condition,  = 2
Product of roots = () =

r
= 22 p ....(i)

q
= 3 p On squaring equation (ii), we get,

Sum of roots = ( + ) = 

....(ii)

q2

92 =

p2

 r  q2  9  2 p  = p2 

 9rp = 2q2

[From equation (i)]

69. c; Given, 3x + 27(3–x) = 12
Let, 3x = y
27
= 12 y 2 y – 12y + 27 = 0 y2 – 9y – 3y + 27 = 0
(y – 3)(y – 9) = 0 y = 3, 9
3x = 3 or 3x = 9 x = 1 or x = 2

 y 







K KUNDAN
1 13
 yy  6







(y2 + 1)6 = 13y
6y2 – 13y + 6 = 0
6y2 – 9y – 4y + 6 = 0
3y(2y – 3) – 2(2y – 3) = 0
(3y – 2)(2y – 3) = 0

y =

2
3
and
3
2

70. a; Let the roots of the given equation x2 – ax + b = 0 be
 and .
 +  = a and  = b
Now, | – | =

(  )2  4

=

a 2  4b

71. b; Given equation is 2x2 – 3x – 4 = 0
To get a reciprocal root, we replace x by

2 x 4


3
1 x 9
 9x = 4 – 4x
When y =

4
x =
13
3 x 9


1 x 4
2
4x = 9 – 9x
When y =

x =

Now, we have,

1
2 
x 

2

1
– 3x  – 4 = 0
 

–4x2 – 3x + 2 = 0
4x2 + 3x – 2 = 0
72. c; Given that, 2x2 + 6x + 5y + 1 = 0 and 2x + y + 3 = 0

....(i)

y  3
2
On putting the value of x in equation (i), we get,

9
13

x =

67. a; Given, 2x2 – 7x + 3 = 0
 2x2– 6x – x + 3 = 0
 2x(x – 3) – 1(x – 3) = 0
 (2x – 1)(x – 3)= 0
When x 

1
.
x

1
,
2

2

 3y 
 3y 
 2 
 6
  5y  1  0
 2 
 2 



9  y 2  6y (18  6y )

 5y  1  0
2
2

 y2 + 10y – 7 = 0

342

SSC Advanced Maths

73. a; If a polynomial equation has rational coefficient and has exactly three real roots, then the degree of the polynomial must be 3.
74. a; Since,  and  are the roots of the equation ax2 + bx
+ c = 0.
 +  = 

b c and  = a a
2

 2   2 
 1
1 


  2  2    2 2 
 

   
=

=

(   )2 (   )2  4



( 2 2 )2

b 2  b 2 4c 

a 2  a 2 a 
 c2

 a2







2

80. c; Given,

=

b2 c4 Let

1
1
1

 x a x b c

(x  b )  ( x  a ) 1
 (x  a )( x  b )  c

(b 2  4ac )

 2cx + (a + b)c = x2 + (a + b)x + ab
 x2 + (a + b – 2c)x + ab – ac – bc = 0
Let the roots of above equation be  and .
Given,  +  = 0
 –(a + b – 2c) = 0
 a + b = 2c
Now,  = ab – ac – bc = ab – (a + b)c
(a  b )
= ab – (a + b)
[From equation (i)]
2

75. b; Given equation is

2x

3x

 1 c (–)    =
c = a a  x
79. c; Given, (x2 – 6x + 45) = 100
 x2 – 6x – 55 = 0
 x2 – 11x + 55x – 55 = 0
x (x – 11) + 5(x – 11) = 0
 (x + 5)(x – 11) = 0
 x = 11, –5

2



78. c; Let the roots of equation ax 2 + bx + c = 0 are –
1
and  .


3x
3

2x
2

2x
a
3x

....(i)

=

2ab  (a 2  b 2  2ab )
(a 2  b 2 )
= 
2
2

81. b; Let the roots of the equation kx 2 – 5x + 6 = 0 be
 and .
5
6
+  = and = k k

2
2
Given,
=
 =


3
3

K KUNDAN
1 3
a   a 2
 2(a2 – 1) = 3a
 2a2 – 3a – 2 = 0
 2a2 – 4a + a – 2 = 0
2a(a – 2) + 1(a – 2) = 0
 (2a + 1)(a – 2) = 0
If a – 2 = 0
2x
= 2
3x
 2x = 4 (3 – x)
6x = 12 x = 2


[from equation (i)]

1
2
But a cannot be negative.
Hence, x = 2 is the required root of the equation.
If 2a + 1 = 0  a = 

76. d; Since,  and  are the roots of the equation x2 – 3x + 2 = 0
 +  = 3 and = 2
.....(i)
Now,  + 1 +  + 1
=  +  + 2 = 3 + 2 = 5
[From equation (i)] and ( + 1)( + 1) =  +  +  + 1
= 2 + 3 + 1 = 6
[From equation (i)]
Thus, required equation is x2 – ( + 1 +  + 1)x + ( + 1)( + 1) = 0
x2 – 5x + 6 = 0
77. c; Since, –1.5 is a root of ax2 + x – 3 = 0
 a(–1.5)2 + (–1.5) –3 = 0
 2.25a – 4.5 = 0
4. 5
a =
= 2
2.25



2
5
2 2
6
+ = and  =
3
k
3
k



5
5
9
=
and 2 =
3
k k =

3
9
and 2 = k k

9
9
= k2 k
 k = 1 and k  0
(  k = 0 does not satisfy the given condition)


82. d; Let  and  be the roots of the equaltion 7x2 + 12x
+ 18 = 0
  +  =

12
18
and  =
7
7

 2 + 2 + 2 =
 2 + 2 =

 2  2

=


144
49

144
36
108
–
= –
49
7
49


108
6
49
= 
18
7
7

343

Quadratic Equations
83. d;

x (x  1)  (m  1) x

(x  1)(m  1) m  m(x2 – x – m – 1) = x(mx – x – m + 1)
 mx2 – mx – m(m + 1) = mx2 – x2 – mx + x
 x2 – x – m(m + 1) = 0
Let the roots be  and .
 +  = 1, and × = –m(m + 1)
2

 =

1 1
   = –m(m + 1)
2 2

 4m2 + 4m + 1 = 0
 (2m + 1)2 = 0
 m= 

1
2

84. a; Since,  and  be the roots of the equation x2 + px + q = 0
 +  = – p and  = q

  p  1 1
 =
Now, ––1 ––1 =     =   q 






 1   1
1
1 and         =
=
 q 
 

Hence, required equation is x2 – (––1 ––1)x + (––1)(––1) = 0

p
1
x   q q
 qx2 – px + 1 = 0
2
 x 

85. c; Since, one root of the equation ax2 + x – 3 = 0 is –1.
 a(–1)2 + (–1) – 3 = 0
 a = 4
 4x2 + x – 3 = 0
Let other root of this equation be .

or y + 2 = 0 y = –2
But, y cannot be negativ e, therefore, it is not permissible. Hence, x = 1 is the required solution.
88. c; Given, 4x – 3.2x + 2 + 32 = 0
 22x – 8.2x – 4.2x + 32 = 0
(2x – 8)(2x – 4) = 0
Either 2x = 8  x = 3 or 2x = 4  x = 2
89. a; Since,  and  be the roots of the equation x2 – x – 1 = 0
 +  = 1 and  = –1
Now, 4 + 4 = (2 + 2)2 – 2() 2
= [( + )2 – 2]2 – 2()2
= (1 + 2)2 – 2 = 9 – 2 = 7
90. d; Let the roots of the quadratic equation be  and .
  +  = 9 and  = 9
Hence, equation is x2 – ( + )x + () = 0
 x2 – 9x + 9 = 0
91. c; We know that the both of the roots of the equation a2x + bx + c = 0 are positive, if

b c  0 and
0
a a Given equation is x 2 – 2 (k – 1)x + (2k + 1) = 0 whose roots are positive. b 2(k  1)

0
=
a
1
 k > 1 c 2(k  1)

0
And
a
1


K KUNDAN
–1 ×  = 

3
3
 =
4
4

86. b; Since, the roots of the equation (a2 + b 2)x 2 –2(ac + bd)x + (c2 + d2) = 0 are equal.
 b2 = 4ac
4(ac + bd)2 = 4(a2 + b2)(c2 + d2)
 a2c2 + b2d2 + 2abcd = a2c2 + a2d2 + b2c2 + b2d2
 (ad – bc)2 = 0
 ad = bc
87. a; Given,

x
–
x 3

x 3
3
=  x 2

1
2
k > 1
Hence, the least value of k in the given answers is 4.
k >

92. b; Here,  +  = 6 and  = 6
 ( + )2 = 62
 2 + 2 + 2 = 36
 2 + 2 = 36 – 2 (6) = 24
(3 + 3) + (2 + 2) + ( + )
= ( + ) (2 + 2 – ) + (2 + 2) + ( + )
= 6 (24 – 6) + (24) + (6)
= 6(18) + 30 = 108 + 30 = 138

93. a; Let the roots of the equation a2b2x2 – (a2 + b2)x + 1
= 0 be  and .
 +  =

Let y =

1
3
x
, then y  y   2 x 3

a 2  b2 a 2b 2

Now,  –  =

and =

1 a 2b 2

(  )2  4

2

 2y – 2 = –3y
2y2 + 3y – 2 = 0
2y2 + 4y – y – 2 = 0
 2y(y + 2) – 1(y + 2) = 0
 (2y – 1)(y + 2) = 0
Either (2y – 1) = 0  y 

x
1
 x 3 4
 4x = x + 3  x = 1


2

=

1
2

 a 2  b2 

  4
 a 2b 2  a 2b 2



(a 2  b 2 )2
 –  =

2 2 2

(a b )



On solving, we get  =
94. b; Given, 2x2 + 3x + c = 0

a 2  b2 a 2b 2

1 b2 and  =

1 a2 344

SSC Advanced Maths
Put x = 0.5, we get,
2(0.5)2 + 3(0.5) + c = 0
 0.5 + 1.5 + c = 0
 c = –2

98. c; Given equation is x2 – 4x + 1 = 0
 x 

95. d; a and c have the same sign opposite to that of b.
96. d; The equation will have equal roots, if b2 – 4ac = 0
(2nc)2 – 4(1 + n2)(c2 – a2) = 0
4n2c2 – 4(c2 + n2c2 – a2 – n2a2) = 0
 –4c2 + 4a2 + 4n2a2 = 0
 c2 = a2(1 + n2)
97. d; Let the roots of the equation x 2 – 2x + 4 = 0 be
 and 
Then,  +  = 2 and  = 4
Taking  as 2 and  as 2 we have,
2 + 2 = 4 and 2 × 2 = +4 × 4 = +16
Therefore, the new equation becomes x2 – 4x + 16 = 0

=

4  16  4  1  1
2 1

42 3
2

= 2  3

When x = 2 +

x3 

1 x3 

3 , then

= 2 3
= (2 +



3

 1 

+ 

2 3 

3 )3 + (2 –

3

3 )3

= 23 + ( 3 )3 + 3 × 2 ×

3 (2 +

– ( 3 )3 – 3 × 2 ×

3 ) + (2)3
3 (2 –

3)

= 8 + 18 + 8 + 18 = 52
Similarly,

when x = 2 –

3
3 , then x 

1 x3 = 52

K KUNDAN