KC Sinha Mathematics Solution Class 10 Chapter 7 Quadratic Equations Exercise 7.4


Exercise 7.1
Exercise 7.2
Exercise 7.3
Exercise 7.4
Exercise 7.5

Exercise 7.4


Question 1 A

Write the discriminate of each of the following quadratic equation:
x2 + 4x + 3 =0
Sol :
d = b2 – 4ac
d = (4)2 – 4 (1) (3)
d = 16 – 12
d = 4

Question 1 B 

Write the discriminate of each of the following quadratic equation:
4x2 + 5x + 7 = 0
Sol :
d = b2 – 4ac
d = (5)2 – 4 (4) (7)
d = 25 – 112
d = –87

Question 1 C

Write the discriminate of each of the following quadratic equation:
2x2 + 4x + 5 = 0
Sol :
d = b2 – 4ac
d = (4)2 – 4 (2) (5)
d = 16 – 40
d = –24

Question 1 D

Write the discriminate of each of the following quadratic equation:
3x2 + 5x + 6 = 0
Sol :
d = b2 – 4ac
d = (5)2 – 4 (3) (6)
d = 25 – 72
d = –47

Question 1 E

Write the discriminate of each of the following quadratic equation:
√3 x2 — 2√2 – 2√3 = 0
Sol :
d = b2– 4ac
d = (–2√2)2 – 4 (√3) (–2√3)
d = 8 + 24
d = 32

Question 2 A

Examine whether the following quadratic equations have real roots or not:
7x2 + 8x — 1 = 0
Sol :
TO CHECK REAL ROOTS, ‘d’ SHOULD BE GREATER THAN 0.
d = b2 – 4ac
d = (8)22 – 4 (7) (–1)
d = 64 – 28
d = 92
Yes, roots are real.

Question 2 B 

Examine whether the following quadratic equations have real roots or not:
2x2 + 3x + 4 = 0
Sol :
TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.
d = b2– 4ac
d = (3)2 – 4(2)(4)
d = 9 – 32
d = – 23
No, roots aren’t real.

Question 2 C

Examine whether the following quadratic equations have real roots or not:
x2 — 12x — 16 = 0
Sol :
TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.
d = b2 – 4ac
d = (–12)2 – 4 (1) (–16)
d = 144 – 64
d = 208
Yes, roots are real .

Question 2 D

Examine whether the following quadratic equations have real roots or not:
x2 + x – 1= 0
Sol :
TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.
d = b2 – 4ac
d = (1)2 – 4 (1) (–1)
d = 1 + 4
d = 5
Yes, roots are real.

Question 2 E

Examine whether the following quadratic equations have real roots or not:
x2 — 10x + 2 = 0
Sol :
TO CHECK REAL ROOTS, ‘d' SHOULD BE GREATER THAN 0.
d = b2 – 4ac
d = (–10)2 – 4 (1) (2)
d = 100 – 8
d = 92
Yes, roots are real.

Question 3 A

Find whether the following quadratic equations have a repeated root :
9x2— 12x + 4 = 0
Sol :
REPEATED ROOTS MEAN d = 0.
d = b2 – 4ac
d = (–12)2 – 4(9)(4)
d = 144 – 144
d = 0
Yes, roots are repeated.

Question 3 B 

Find whether the following quadratic equations have a repeated root :
y2 — 6y + 6 = 0
Sol :
REPEATED ROOTS MEAN d = 0.
d = b2 – 4ac
d = (–6)2 – 4(1)(6)
d = 36 – 24
d = 12
 roots are not repeated.

Question 3 C

Find whether the following quadratic equations have a repeated root :
9x2 + 4x + 6 = 0
Sol :
REPEATED ROOTS MEAN d = 0.
d = b2 – 4ac
d = (4)2 – 4 (9) (6)
d = 16 – 216
d = –200
 roots are not repeated.

Question 3 D

Find whether the following quadratic equations have a repeated root :
16y2— 40y + 25 = 0
Sol :
REPEATED ROOTS MEAN d = 0 .
d = b2 – 4ac
d = (–40)2 – 4 (16) (25)
d = 1600 – 1600
d = 0
 roots are repeated.

Question 3 E

Find whether the following quadratic equations have a repeated root :
x2 + 6x + 9 = 0
Sol :
REPEATED ROOTS MEAN d = 0 .
d = b2 – 4ac
d = (6)2 – 4(1)(9)
d = 36 – 36
d = 0
 roots are repeated.

Question 4 A

Comment upon the nature of roots of the following equations:
4x2 + 7x + 2 = 0
Sol :
d = b2 – 4ac
d = (7)2 – 4 (4) (2)
d = 49 – 32
d = 17
Since, d>0, roots are unique and real.

Question 4 B 

Comment upon the nature of roots of the following equations:
x2 + 10x + 39 = 0
Sol :
d = b2 – 4ac
d = (10)2 – 4 (1) (39)
d = 100 – 156
d = –56
Since, d < 0, no real roots exists.

Question 5 A

Without solving, determine whether the following equations have real roots or not. If yes, find them:
2x2— 4x + 3 = 0
Sol :
d = b2 – 4ac
d = (–4)2 – 4 (2) (3)
d = 16 – 24
d = –8
Since, d < 0, no real roots exist for the given equation.

Question 5 B

Without solving, determine whether the following equations have real roots or not. If yes, find them:
$\mathrm{y}^{2}-\frac{2}{3} \mathrm{y}+\frac{1}{9}=0$
Sol :
d = b2 – 4ac
$\mathrm{d}=\left(\frac{-2}{3}\right)^{2}-4(1)\left(\frac{1}{9}\right)$
$d=\frac{4}{9}-\frac{4}{9}$
d = 0
Since, d = 0, roots are real and equal for the given equation.
$x=\frac{-b \pm \sqrt{d}}{2 a}$
$\mathrm{x}=\frac{-\left(-\frac{2}{3}\right) \pm \sqrt{0}}{2 \times 1}$
$x=\frac{2}{3} \times \frac{1}{2}$
$x=\frac{1}{3}$

Question 6 A

Without finding the roots, comment upon the nature of roots of each of the following quadratic equations:
2x2— 6x + 3 = 0
Sol :
d = b2 – 4ac
d = (–6)2 – 4 (2) (3)
d = 36 – 24
d = 12
 Roots are real and unique.

Question 6 B

Without finding the roots, comment upon the nature of roots of each of the following quadratic equations:
2x2 — 5x — 3 = 0
Sol :
d = b2 – 4ac
d = (–5)2 – 4 (2) (–3)
d = 25 + 24
d = 49
 Roots are real and unique.

Question 7 A 

Find the value of k for which the quadratic equation
4x2 — 2 (k + 1) x + (k + 4) = 0 has equal roots.
Sol :
Since roots are equal
 d=0 ….(1)
4x2 — 2 (k + 1) x + (k + 4) = 0
d = b2 – 4ac
d = (–2 (k + 1)2 – 4 (4) (k + 4)
d = (– 2k – 2)2 – 16k – 64
d = 4k2 + 4 + 8k – 16k – 64
( (a – b)2 = a2 + b2 – 2ab)
d = 4k2 – 8k – 60
From (1), d = 0
 Equation will be:
4k2 – 8k – 60 = 0
Dividing by 4
k2 – 2k – 15 = 0
4k2 – 5k + 3k – 15 = 0
k (k – 5) + 3(k – 5) = 0
(k – 5) (k + 3) = 0
K – 5 = 0 k + 3 = 0
K = 5 k = –3

Question 7 B

Find the value of k, so that the quadratic equation
(k + 1) x2 – 2 (k — 1) x + 1 = 0 has equal roots.
Sol :
Since roots are equal
 d=0 ….(1)
(k + 1)x2 — 2 (k – 1) x + 1 = 0
d = b2 – 4ac
d = (–2(k–1))2– 4(k+1)(1)
d = (–2k+2)2 – 4k – 4
d=4k2 + 4 – 8k – 4k – 4
( (a + b)2 = a2 + b2 + 2ab)
d = 4k2 – 12k
From (1), d = 0
 Equation will be:
0 = 4k2 – 12k
4k2 = 12k
$\mathrm{k}^{2}=\frac{12}{4} \mathrm{k}$
k2 = 3k
k2 – 3k = 0
k(k – 3) = 0
k = 0 or k – 3 = 0
k = 3
 Values of k are 0, 3.

Question 8 A

For what values of k, does the following quadratic equation has equal roots.
9x2 + 8kx + 16 = 0
Sol :
Since roots are equal
 d=0 (1)
9x2 + 8kx + 16 = 0
d = b2 – 4ac
d = (8k)2 – 4 (9) (16)
From (1), d = 0
 Equation will be:
0 = 64k2 – 576
$\mathrm{k}^{2}=\frac{576}{64}$
k2 = 9
k = ±√9
k = 3, –3
 values of k are –3, 3 .

Question 8 B

(k + 4)x2 + (k + 1)x + 1 = 0
Sol :
Since roots are equal
 d=0 (1)
(k + 4)x2 + (k + 1)x + 1 = 0
d=b2–4ac
d = (k – 1)2– 4 (k + 4) (1)
d = (–2k + 2)2 – 4k – 4
d = k2 + 1+ 2k – 4k – 16
From (1), d = 0
 Equation will be:
0 = k2 + 1 + 2k – 4k – 16
k2 – 2k – 15 = 0
k2 – 5k + 3k – 15 = 0
k(k – 5) + 3 (k – 5) = 0
(k – 5) (k + 3) = 0
K – 5 = 0 k + 3 = 0
k = 5 k = –3
 Values of k are –3, 5.

Question 8 C

k2x2 — 2(2k — 1)x + 4 = 0
Sol :
Since roots are equal
 d=0 (1)
k2x2 — 2(2k — 1)x + 4 = 0
d = b2 – 4ac
d = (–2(k–1))2 – 4 (k2) (4)
d = (–2k+2)2 – 4k – 4
d = (–4k+2)2 – 16k2
( (a – b)2 = a2 + b2 – 2ab)
d = 16k2 – 16k + 4 – 16k2
d = –16k + 4
From (1), d = 0
 Equation will be:
0 = –16k + 4
16k = 4


 Values of k are is .

Question 9 

If the roots of the equation (a — b)x2 + (b — c) x + (c — a) = 0 are equal, prove that 2a = b + c.
Sol :
Since roots are equal
 d=0 (1)
(a — b)x2 + (b — c) x + (c — a) = 0
d = b2 – 4ac
d = (b–c)2 – 4 (a–b) (c–a)
d = b2 + c2 – 2bc –4 [a (c – a) – b (c – a)]
d = b2 + c2 – 2bc – 4 [ac – a2 – bc + ba]
From (1), d = 0
 Equation will be:
0 = b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ba
b2 + c2 – (2a)2 2bc + 2c (–2a) + 2(–2a)b = 0
(b + c – 2a)2 = 0
(b + c – 2a) = 0
b + c = 2a
Hence proved.

Question 10

If — 5 is a root of the quadratic equation 2x2 + 2px — 15 = 0 and the quadratic equation p (x2 + x) + k = 0 has equal roots, find the value of k.
Sol :
2x2 + 2px — 15 = 0
Put x = –5
2(–5)2 + 2p(–5) — 15 = 0
50 – 10p – 15 = 0
35 = 10p
p = 3.5
Equation p (x2 + x) + k = 0 has equal roots i.e. d = 0
p (x2 + x) + k = 0
p = 3.5
3.5 (x2 + x) + k = 0
3.5x2 + 3.5x + k = 0
d = b2 – 4ac
d = (3.5)2 – 4(3.5)(k)
d = 12.25 – 14k
Putting d = 0
 Equation will be:
0 = 12.25 – 14k
14k = 12.25
$\mathrm{k}=\frac{12.25}{14}$
k = 0.875

Question 11 A

Find the values of k, for which the given equation has real roots:
2x2 — 10x + k = 0
Sol :
Roots are equal
 d = 0
d = b2 – 4ac
d = (–10)2 – 4 (2) (k)
d = 100 – 8k
Put d = 0
0 = 100 – 8k
8k = 100
$\mathrm{k}=\frac{100}{8}$
$\mathrm{k}=\frac{25}{2}$

Question 11 B

Find the values of k, for which the given equation has real roots:
kx2 — 6x — 2 = 0
Sol :
Roots are equal
 d = 0
d = b2 – 4ac
d = (–6)2 – 4 (–2) (k)
d = 36 + 8k
Put d = 0
0 = 36 + 8k
–8k = 36
$\mathrm{k}=\frac{-36}{8}$
$\mathrm{k}=\frac{-9}{2}$

Question 11 C

Find the values of k, for which the given equation has real roots:
kx2+ 4x + 1 = 0
Sol :
Roots are equal
 d = 0
d = b2 – 4ac
d = (4)2 – 4 (1) (k)
d = 16 – 4k
Put d = 0
0 = 16 – 4k
4k = 16
k = 4

Question 11 D

Find the values of k, for which the given equation has real roots:
kx2 – 2√5x + 4 = 0
Sol :
Roots are equal
 d = 0
d = b2 – 4ac
d = (–2√5)2 – 4 (k) (4)
d = 20 – 16k
Put d = 0
0 = 20 – 16k
16k = 20
$\mathrm{k}=\frac{20}{16}$
$\mathrm{k}=\frac{5}{4}$

Question 11 E

Find the values of k, for which the given equation has real roots:
x2 + k(4x + k — 1) + 2 = 0
Sol :
Roots are equal
 d = 0
d = b2 – 4ac
d = (4k)2 – 4 (k2 – k + 2) (1)
d = 16k2 – 4k2 + 4k – 8
Put d = 0
0 = 16k2 – 4k2 + 4k – 8
12k2 + 4k2 + 4k – 8 = 0 (divide by 4)
3k2 + k – 2 = 0
3k2 + 3k – 2k – 2 = 0
3k (k + 1) – 2k (k + 1) = 0
(3k–2k)(k+1) = 0
(3k–2k)=0 or (k+1) = 0
k = 0 k = –1

Question 12

Prove that the equation x2(a2 + b2) + 2x(ac + bd) + (c2 + d2)= 0 has no real root, if ad ≠ bc.
Sol :
x2(a2 + b2) + 2x(ac + bd) + (c2 + d2)= 0
d = b2 – 4ac
d = (2ac + 2bd)2 – 4 (a2 + b2) (c2 + d2)
d = 4a2c2 + 4b2d2 + 8abcd – 4 [a2 (c2 + d2) + b2 (c2+d2)]
d = 4a2c2 + 4b2d2 + 8abcd – 4 [a2c2 + a2d2 + b2c2 + b2d2]
d = 4a2c2 + 4b2d2 + 8abcd – 4a2c2 – 4a2d2 – 4b2c2 – 4b2 d2
d = 8abcd – 4a2d2 – 4b2c2
d = 8abcd – 4(a2d2 + b2 c2)
d = –4 (a2 d2 + b2c2 – 2abcd)
d = –4 [(ad + bc)2]
For ad ≠ bc
d= –4 × [value of (ad + bc)2]
 d is always negative
So, d < 0
The given equation has no real roots.


S.no Chapters Links
1 Real numbers Exercise 1.1
Exercise 1.2
Exercise 1.3
Exercise 1.4
2 Polynomials Exercise 2.1
Exercise 2.2
Exercise 2.3
3 Pairs of Linear Equations in Two Variables Exercise 3.1
Exercise 3.2
Exercise 3.3
Exercise 3.4
Exercise 3.5
4 Trigonometric Ratios and Identities Exercise 4.1
Exercise 4.2
Exercise 4.3
Exercise 4.4
5 Triangles Exercise 5.1
Exercise 5.2
Exercise 5.3
Exercise 5.4
Exercise 5.5
6 Statistics Exercise 6.1
Exercise 6.2
Exercise 6.3
Exercise 6.4
7 Quadratic Equations Exercise 7.1
Exercise 7.2
Exercise 7.3
Exercise 7.4
Exercise 7.5
8 Arithmetic Progressions (AP) Exercise 8.1
Exercise 8.2
Exercise 8.3
Exercise 8.4
9 Some Applications of Trigonometry: Height and Distances Exercise 9.1
10 Coordinates Geometry Exercise 10.1
Exercise 10.2
Exercise 10.3
Exercise 10.4
11 Circles Exercise 11.1
Exercise 11.2
12 Constructions Exercise 12.1
13 Area related to Circles Exercise 13.1
14 Surface Area and Volumes Exercise 14.1
Exercise 14.2
Exercise 14.3
Exercise 14.4
15 Probability Exercise 15.1

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