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# Class 10 RD SHARMA Solutions Maths Chapter 4 - Quadratic Equations

TopperLearning’s class 10 R.D Sharma solutions in Quadratic Equations are the best when the primary focus is scoring maximum grades. The in-house subject matter experts with years of experience and subjective knowledge have developed the chapter-wise R.D Sharma Maths solutions. Here, multiple illustrative examples and chronologically step-by-step solutions make learning easy. That way, you can aim to score high marks in the upcoming CBSE Class 10 Maths board exam.

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## Quadratic Equations Exercise Ex. 4.1

### Solution 2 (v)

= 2x- x + 9 – x+ 4x + 3

= x2 - 5x + 6 = 0

Here, LHS = x2 - 5x + 6 and RHS = 0

Substituting x = 2 and x = 3

= x2 - 5x + 6

= (2)2 – 5(2) + 6

=10-10

=0

= RHS

= x2 - 5x + 6

= (3)2 – 5(3) + 6

= 9 - 15 + 6

=15 - 15

=0

= RHS

x = 2 and x = 3 both are the solutions of the given quadratic equation.

## Quadratic Equations Exercise Ex. 4.4

### Solution 6

Given equation is x2 - 8x + 18 = 0

x2 - 2 × x × 4 + + 42 - 42 + 18 = 0

(x - 4)2 - 16 + 18 = 0

(x - 4)2 = 16 - 18

(x - 4)2 = -2

Taking square root on both the sides, we get

Therefore, real roots does not exist.

## Quadratic Equations Exercise Ex. 4.5

### Solution 1 (vii)

x2 + 2 × x × 5 + 52 = 10x - 6

x2 + 10x + 25 = 10x - 6

x2 + 31 = 0

Here, a = 1, b = 0 and c = 31

Therefore, the discriminant is

D = b2 - 4ac

= 0 - 4 × 1 × 31

= -124

## Quadratic Equations Exercise Ex. 4.6

### Solution 1 (vi)

Here,

Therefore, we have

As D = 0, roots of the given equation are real and equal.

### Solution 2(xviii)

4x2 - 2 (k + 1)x + (k + 1) = 0 Comparing with ax2 + bx + c = 0, we get a = 4, b = -2(k + 1), c = k + 1 According to the question, roots are real and equal. Hence, b2 - 4ac = 0

### Solution 4(iv)

x2 + k(2x + k - 1) + 2 = 0 x2 + 2kx + k(k - 1) + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 2k, c = k(k - 1) + 2 According to the question, roots are real and equal. Hence, b2 - 4ac = 0

### Solution 5(vi)

4x2 + kx + 3 = 0 Comparing with ax2 + bx + c = 0, we get a = 4, b = k, c = 3 According to the question, roots are real and equal. Hence, b2 - 4ac = 0

### Solution 9 (ii)

Given quadratic equation is x2 + kx + 16 = 0

As it has equal roots, the discriminant will be 0.

Here, a = 1, b = k, c = 16

Therefore, D = k2 - 4(1)(16) = 0

i.e. k2 - 64 = 0

i.e. k = ± 8

When k = 8, the equation becomes x2 + 8x + 16 = 0

or x2 - 8x + 16 = 0

As D = 0, roots of the given equation are real and equal.

## Quadratic Equations Exercise Ex. 4.7

### Solution 24

Let x be the natural number.

As per the question, we have

Therefore, x = 8 as x is a natural number.

Hence, the required natural number is 8.

## Quadratic Equations Exercise Ex. 4.8

### Solution 10

Concept Insight: Use the relation s =d/t to crack this question and remember here distance is constant so speed and time will vary inversely.

### Solution 15

Let the speed of a car be x km/hr. According to the question, time is hr. Distance = Speed × Time 2592 = x = 72 km/hr Hence, the time taken by a car to cover a distance of 2592 km is 36 hrs.

### Solution 16

Let x km/hr be the speed of the stream.

Therefore, we have

Downstream speed = (9 + x) km/hr

Upstream speed = (9 - x) km/hr

Distance covered downstream = distance covered upstream

Total time taken = 3 hours 45 minutes = hours

Therefore, x = 3 as the speed can't be negative.

Hence, the speed of the motor boat is 3 km/hr.

## Quadratic Equations Exercise Ex. 4.12

### Solution 5

Let us assume that the larger pipe takes 'x' hours to fill the pool.

So, as per the question, the smaller pipe takes 'x + 10' hours to fill the same pool.

### Solution 6

Let the tap with smaller diameter takes x hours to completely fill the tank.

So, the other tap takes (x - 2) hours to fill the tank completely.

Total time taken to fill the tank  hours

As per the question, we have

When x = 5, then (x - 2) = 3

When  which can't be possible as the time becomes negative.

Hence, the smaller diameter tap fills in 5 hours and the larger diameter tap fills in 3 hours.

## Quadratic Equations Exercise Ex. 4.13

### Solution 1

We know for the quadratic equation

ax2 + bx + c = 0

condition for roots to be real and distinct is

D = b2 - 4ac > 0       ..........(1)

for the given question

x+ 4x + k = 0

a = 1, b = 4, c = k

from (1)

16 - 4k > 0

k < 4

So, the correct option is (a).

### Solution 2

For the equation x2 - ax + 1 = 0 has two distinct roots, condition is

(-a)- 4 (1) (1) > 0

a- 4 > 0

a> 4

|a| > 2

So, the correct option is (c).

### Solution 5

ax + bx + c = 0

having two distinct roots, condition is

b2 - 4ac > 0

For the equation ax2 + 2x + a = 0 to have two distinct roots,

(2)2 - 4 (a) (a) > 0

4 - 4a2 > 0

4(1 - a2) > 0

1 - a2 > 0 since 4 > 0

that is, a2 - 1 < 0

Hence -1 < a < 1, only integral solution possible is a = 0

So, the correct option is (b).

### Solution 6

ax2 + bx + c = 0

having real roots, condition is

b2 - 4ac ≥ 0     .......(1)

According to question

x+ kx + 64 = 0 have real root if

k- 4 × 64 ≥ 0

k2 ≥ 256

|k|≥ 16             ........(2)

Also, x2 - 8x + k = 0 has real roots if

64 - 4k ≥ 0

k ≤ 16    .........(3)

from (2), (3) the only positive solution for k is

k = 16

So, the correct option is (d).

### Solution 8

It is given that 2 is a root of equation x + bx + 12 = 0

Hence

(2)+ b(2) + 12 = 0

4 + 2b + 12 = 0

2b + 16 = 0

b = -8        .........(1)

It is also given that x2 + bx + q = 0 has equal root

so, b2 - 4(q) = 0         ........(2)

from (1) & (2)

(-8)- 4q =0

q = 16

So, the correct option is (c).

### Solution 9

ax2 + bx + c = 0

Having equal roots, the condition is

b - 4ac = 0

For the equation

(a2 + b2) x2 - 2 (ac + bd)x + c2 + d= 0

to have equal roots, we have

(-2(ac + bd))- 4 (c+ d2) (a2 + b2) = 0

4 (ac + bd)- 4 (a2c+ b2c2 + d2a2 + b2d2) = 0

(a2c+ b2d2 + 2abcd) - (a2c+ b2c2 + a2d2+ b2d2) = 0

2abcd - b2c2 - a2d2 = 0

b2c2 - a2d- 2abcd = 0

So, the correct option is (b).

### Solution 10

ax+ bx + c = 0

having equal roots, condition is

b2 - 4ac = 0

According to question, quadratic equation is

(a2 + b2)x2 - 2b(a + c)x + b2 + c= 0

having equal roots, so

(2b(a + c))2 - 4(a2 + b2) (b2 + c2) = 0

4b2 (a + c)2 - 4(a2b2 + a2c2 + b4 + b2c2) = 0

b2(a2 + c2 + 2ac) - (a2b2 + a2c+ b2c2 + b4) = 0

a2b2 + b2c2 + 2acb2 - a2b2 - a2c- b2c2 - b4 = 0

2acb2 - a2c2 - b4 = 0

a2c2 + b4 - 2acb2 = 0

(ac - b2)2 = 0

ac = b2

So, the correct option is (b).

### Solution 11

For any quadratic equation ax2 + bx + c = 0 having no real roots, condition is

b2 - 4ac < 0

For the equation, x2 - bx + 1 = 0 having no real roots

b2 - 4 < 0

b2 < 4

-2 < b < 2

So, the correct option is (b).

### Solution 13

If p, q are the roots of equation x- px + q = 0, then p and q satisfies the equation

Hence

(p)- p(p) + q = 0

p- p+ q = 0

q = 0

and (q)- p(q) + q = 0

q- p(q) + q = 0

0 = 0

p can take any value

p = -2 and q = 0

So, the correct option is (c).

### Solution 14

For the ax+ bx + 1 = 0 having real roots condition is

b- 4(a) (1) ≥ 0

b2 ≥ 4a

For a = 1

b2 ≥ 4

b ≥ 2

b can take value 2, 3, 4

Here, 3 possible solutions are possible            .......(1)

For a = 2

b2 ≥ 8

Here, b can take value 3, 4            ......(2)

Here, 2 solutions are possible

For a = 3

b≥ 12

possible value of b is 4

Hence, only 1 possible solution      ......(3)

For a = 4

b2 ≥ 16

possible value of b is 4

Hence, only 1 possible solution    .......(4)

from (1), (2), (3), (4)

Total possible solutions are 7

So, the correct option is (b).

### Solution 15

Any quadratic equation having roots 0 or 1 are only possible quadratic equation because on squaring 0 or 1, it remains same.

Hence, 2 solutions are possible, one having roots 1 and 1, while the other having roots 0 and 1.

So, the correct option is (c).

### Solution 16

If any quadratic equation ax+ bx + c has no real roots then b- 4ac < 0    ......(1)

According to the question, the equation is

(a+ b2) x+ 2(ac + bd) x + c2 + d= 0

from (1)

4(ac + bd)- 4(a2 + b2) (c2 + d2) < 0

a2c2 + b2d2 + 2abcd - (a2c2 + a2d2 + b2c2 + b2d2) < 0

a2c2 + b2d2 + 2abcd - a2c2 - a2d2 - b2c2 - b2d2 < 0

2abcd - a2d2 - b2c2 < 0

For this condition to be true ad ≠ bc

So, the correct option is (d).

### Solution 18

It is given that x = 1 is root of equation ax2 + ax + 2 = 0

Hence, a(1)2 + a(1) + 2 = 0

2a + 2 = 0

a = -1         ......(1)

It is given that x = 1 is also root of x2 + x + b = 0

Hence, (1)+ (1) + b = 0

b = -2     ......(2)

from (1) & (2)

ab = (-1) (-2)

ab = 2

So, the correct option is (b).

### Solution 23

Given, 2 is a root of equation x+ ax + 12 = 0

so (2)+ a(2) + 12 = 0

4 + 2a + 12 = 0

a = -8     .....(1)

Given x+ ax + q = 0 has equal roots so

a- 4q = 0   ......(2)

from (1) & (2)

(-8)- 4q = 0

4q = 64

q = 16

So, the correct option is (d).

### Solution 25

If a and b are roots of the equation x+ ax + b = 0

Then, sum of roots = -a

a + b = -a

2a + b = 0          .......(1)

product of roots = b

ab = b

ab - b = 0

b(a - 1) = 0         ........(2)

from (1) and (2)

-2a(a - 1) = 0...(From (1), we have b = -2a)

a(a - 1) = 0

a = 0 or a = 1

if a = 0  b = 0

if a = 1  b = -2

Now a and b can't be zero at same time, so correct solution is

a = 1 and b = -2

a + b = -1

So, the correct option is (d).

### Solution 26

Given sum of roots is zero and one root is 2.

So the other root must be -2

so any quadratic equation having root 2 and -2 is

(x - 2) (x - (-2)) = 0

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

x- 4 = 0

So, the correct option is (b).

### Solution 29

x+ ax + 3 = 0

product of roots = 3

One root is 1. Hence other root is 3.

So, the correct option is (a).

### Solution 32

Any quadratic equation, ax+ bx + c = 0 has real and equal roots if b- 4ac = 0

For the question, equation is 16x+ 4kx + 9 = 0,

(4k)- 4 × 16 × 9 = 0

16k- 36 × 16 = 0

k2 - 36 = 0

k= 36

k = ± 6

So, the correct option is (c).