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

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:

y2−23y+19=0

Sol :
d = b² – 4ac
d=(−23)2−4(1)(19)
d=49−49
d = 0
Since, d = 0, roots are real and equal for the given equation.
x=−b±d√2a
x=−(−23)±0√2×1
x=23×12
x=13

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
k2=124k
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
k2=57664
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
k=12.2514
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

k=1008

k=252

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

k=−368

k=−92

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

k=2016

k=54

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.