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

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## Quadratic Equations Exercise Ex. 4.1

### Solution 1 (i)

### Solution 1 (ii)

### Solution 1 (iii)

### Solution 1 (iv)

### Solution 1 (v)

### Solution 1(vi)

### Solution 1 (vii)

### Solution 1 (viii)

### Solution 1 (ix)

### Solution 1 (x)

### Solution 1 (xi)

### Solution 1 (xii)

### Solution 1 (xiii)

### Solution 1 (xiv)

### Solution 1 (xv)

### Solution 2 (i)

### Solution 2 (ii)

### Solution 2 (iii)

### Solution 2 (iv)

### Solution 2 (v)

= 2x^{2 }- x + 9 – x^{2 }+ 4x + 3

= x^{2} - 5x + 6 = 0

Here, LHS = x^{2} - 5x + 6 and RHS = 0

Substituting x = 2 and x = 3

= x^{2} - 5x + 6

= (2)^{2} – 5(2) + 6

=10-10

=0

= RHS

= x^{2} - 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.

### Solution 2 (vi)

### Solution 2 (vii)

### Solution 3 (i)

### Solution 3 (ii)

### Solution 3 (iii)

### Solution 3 (iv)

### Solution 4

### Solution 5

## Quadratic Equations Exercise Ex. 4.2

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

## Quadratic Equations Exercise Ex. 4.3

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

### Solution 22

### Solution 23

### Solution 24

### Solution 25

### Solution 26

### Solution 27

### Solution 28

### Solution 29

### Solution 30

### Solution 31

### Solution 32

### Solution 33

### Solution 34

### Solution 35

### Solution 36

### Solution 37

### Solution 38

### Solution 39

### Solution 40

### Solution 41

### Solution 42

### Solution 43

### Solution 44

### Solution 45

### Solution 46

### Solution 47

### Solution 48

### Solution 49

### Solution 50

### Solution 51

### Solution 52

### Solution 53

### Solution 54

### Solution 55

### Solution 56

### Solution 57

### Solution 58

### Solution 59

### Solution 60

### Solution 61

### Solution 62

## Quadratic Equations Exercise Ex. 4.4

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

Given equation is x^{2} - 8x + 18 = 0

x^{2} - 2 × x × 4 + + 4^{2} - 4^{2}
+ 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.

### Solution 7

### Solution 8

### Solution 9

### Solution 10

## Quadratic Equations Exercise Ex. 4.5

### Solution 1 (i)

### Solution 1 (ii)

### Solution 1(iii)

### Solution 1 (iv)

### Solution 1(v)

### Solution 1(vi)

### Solution 1 (vii)

x^{2} + 2 × x × 5 + 5^{2} = 10x - 6

x^{2} + 10x + 25 = 10x - 6

x^{2} + 31 = 0

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

Therefore, the discriminant is

D = b^{2} - 4ac

= 0 - 4 × 1 × 31

= -124

### Solution 2 (i)

### Solution 2 (ii)

### Solution 2 (iii)

### Solution 2 (iv)

### Solution 2 (v)

### Solution 2 (vi)

### Solution 2 (vii)

### Solution 2 (viii)

### Solution 2 (ix)

### Solution 2(x)

### Solution 2(xi)

### Solution 2(xii)

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 3(iv)

### Solution 3(v)

## Quadratic Equations Exercise Ex. 4.6

### Solution 1(i)

### Solution 1(ii)

### Solution 1(iii)

### Solution 1(iv)

### Solution 1(v)

### Solution 1 (vi)

Given quadratic equation is

Here,

Therefore, we have

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

### Solution 2(i)

### Solution 2(ii)

### Solution 2(iii)

### Solution 2(iv)

### Solution 2(v)

### Solution 2(vi)

### Solution 2(vii)

### Solution 2(viii)

### Solution 2(ix)

### Solution 2(x)

### Solution 2(xi)

### Solution 2(xii)

### Solution 2(xiii)

### Solution 2(xiv)

### Solution 2(xv)

### Solution 2(xvi)

### Solution 2(xvii)

### Solution 2(xviii)

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

### Solution 3(i)

### Solution 3(ii)

### Solution 3(iii)

### Solution 3(iv)

### Solution 3(v)

### Solution 4(i)

### Solution 4(ii)

### Solution 4(iii)

### Solution 4(iv)

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

### Solution 5(i)

### Solution 5(ii)

### Solution 5(iii)

### Solution 5(iv)

### Solution 5(v)

### Solution 5(vi)

4x^{2}
+ kx + 3 = 0
Comparing with ax^{2} + bx + c = 0, we get
a = 4, b = k, c = 3
According to the question, roots are real and equal.
Hence, b^{2} - 4ac = 0

### Solution 6 (i)

### Solution 6 (ii)

### Solution 7

### Solution 8

### Solution 9(i)

### Solution 9 (ii)

Given quadratic
equation is x^{2} + kx + 16 = 0

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

Here, a = 1, b = k, c = 16

Therefore, D = k^{2}
- 4(1)(16) = 0

i.e. k^{2}
- 64 = 0

i.e. k = ± 8

When k = 8, the
equation becomes x^{2} + 8x + 16 = 0

or x^{2} - 8x + 16 = 0

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

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15(i)

### Solution 15(ii)

### Solution 15(iii)

### Solution 15(iv)

### Solution 16(i)

### Solution 16(ii)

### Solution 16(iii)

### Solution 16(iv)

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

### Solution 22

### Solution 23

### Solution 24

### Solution 25

## Quadratic Equations Exercise Ex. 4.7

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

### Solution 22

### Solution 23

### 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.

### Solution 25

### Solution 26

### Solution 27

### Solution 28

### Solution 29

### Solution 30

### Solution 31

### Solution 32

### Solution 33

### Solution 34

### Solution 35

### Solution 36

### Solution 37

### Solution 38

### Solution 39

### Solution 40

## Quadratic Equations Exercise Ex. 4.8

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### 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 11

### Solution 12

### Solution 13

### Solution 14

### 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. 4.9

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

## Quadratic Equations Exercise Ex. 4.10

### Solution 1

### Solution 2

### Solution 3

### Solution 4

## Quadratic Equations Exercise Ex. 4.11

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

## Quadratic Equations Exercise Ex. 4.12

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### 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

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

## Quadratic Equations Exercise 4.82

### Solution 1

We know for the quadratic equation

ax^{2} + bx + c = 0

condition for roots to be real and distinct is

D = b^{2} - 4ac > 0 ..........(1)

for the given question

x^{2 }+ 4x + k = 0

a = 1, b = 4, c = k

from (1)

16 - 4k > 0

k < 4

So, the correct option is (a).

## Quadratic Equations Exercise 4.83

### Solution 2

For the equation x^{2} - ax + 1 = 0 has two distinct roots, condition is

(-a)^{2 }- 4 (1) (1) > 0

a^{2 }- 4 > 0

a^{2 }> 4

|a| > 2

So, the correct option is (c).

### Solution 3

### Solution 4

### Solution 5

For any quadratic equation

ax^{2 } + bx + c = 0

having two distinct roots, condition is

b^{2} - 4ac > 0

For the equation ax^{2} + 2x + a = 0 to have two distinct roots,

(2)^{2} - 4 (a) (a) > 0

4 - 4a^{2} > 0

4(1 - a^{2}) > 0

1 - a^{2} > 0 since 4 > 0

that is, a^{2} - 1 < 0

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

So, the correct option is (b).

### Solution 6

For any quadratic equation

ax^{2} + bx + c = 0

having real roots, condition is

b^{2} - 4ac ≥ 0 .......(1)

According to question

x^{2 }+ kx + 64 = 0 have real root if

k^{2 }- 4 × 64 ≥ 0

k^{2} ≥ 256

|k|≥ 16 ........(2)

Also, x^{2} - 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 7

### Solution 8

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

Hence

(2)^{2 }+ b(2) + 12 = 0

4 + 2b + 12 = 0

2b + 16 = 0

b = -8 .........(1)

It is also given that x^{2} + bx + q = 0 has equal root

so, b^{2} - 4(q) = 0 ........(2)

from (1) & (2)

(-8)^{2 }- 4q =0

q = 16

So, the correct option is (c).

### Solution 9

For any quadratic equation

ax^{2} + bx + c = 0

Having equal roots, the condition is

b^{2 } - 4ac = 0

For the equation

(a^{2} + b^{2}) x^{2} - 2 (ac + bd)x + c^{2} + d^{2 }= 0

to have equal roots, we have

(-2(ac + bd))^{2 }- 4 (c^{2 }+ d^{2}) (a^{2 }+ b^{2}) = 0

4 (ac + bd)^{2 }- 4 (a^{2}c^{2 }+ b^{2}c^{2} + d^{2}a^{2 }+ b^{2}d^{2}) = 0

(a^{2}c^{2 }+ b^{2}d^{2} + 2abcd) - (a^{2}c^{2 }+ b^{2}c^{2} + a^{2}d^{2}+ b^{2}d^{2}) = 0

2abcd - b^{2}c^{2} - a^{2}d^{2} = 0

b^{2}c^{2} - a^{2}d^{2 }- 2abcd = 0

(bc - ad)^{2 }= 0

bc = ad

So, the correct option is (b).

### Solution 10

For any quadratic equation

ax^{2 }+ bx + c = 0

having equal roots, condition is

b^{2} - 4ac = 0

According to question, quadratic equation is

(a^{2} + b^{2})x^{2} - 2b(a + c)x + b^{2} + c^{2 }= 0

having equal roots, so

(2b(a + c))^{2} - 4(a^{2} + b^{2}) (b^{2} + c^{2}) = 0

4b^{2} (a + c)^{2} - 4(a^{2}b^{2} + a^{2}c^{2 }+ b^{4} + b^{2}c^{2}) = 0

b^{2}(a^{2} + c^{2} + 2ac) - (a^{2}b^{2} + a^{2}c^{2 }+ b^{2}c^{2} + b^{4}) = 0

a^{2}b^{2} + b^{2}c^{2} + 2acb^{2} - a^{2}b^{2} - a^{2}c^{2 }- b^{2}c^{2} - b^{4} = 0

2acb^{2} - a^{2}c^{2} - b^{4} = 0

a^{2}c^{2} + b^{4} - 2acb^{2} = 0

(ac - b^{2})^{2} = 0

ac = b^{2}

So, the correct option is (b).

### Solution 11

For any quadratic equation ax^{2} + bx + c = 0 having no real roots, condition is

b^{2} - 4ac < 0

For the equation, x^{2} - bx + 1 = 0 having no real roots

b^{2} - 4 < 0

b^{2} < 4

-2 < b < 2

So, the correct option is (b).

### Solution 12

### Solution 13

If p, q are the roots of equation x^{2 }- px + q = 0, then p and q satisfies the equation

Hence

(p)^{2 }- p(p) + q = 0

p^{2 }- p^{2 }+ q = 0

q = 0

and (q)^{2 }- p(q) + q = 0

q^{2 }- 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^{2 }+ bx + 1 = 0 having real roots condition is

b^{2 }- 4(a) (1) ≥ 0

b^{2} ≥ 4a

For a = 1

b^{2} ≥ 4

b ≥ 2

b can take value 2, 3, 4

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

For a = 2

b^{2} ≥ 8

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

Here, 2 solutions are possible

For a = 3

b^{2 }≥ 12

possible value of b is 4

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

For a = 4

b^{2} ≥ 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).

## Quadratic Equations Exercise 4.84

### Solution 16

If any quadratic equation ax^{2 }+ bx + c has no real roots then b^{2 }- 4ac < 0 ......(1)

According to the question, the equation is

(a^{2 }+ b^{2}) x^{2 }+ 2(ac + bd) x + c^{2} + d^{2 }= 0

from (1)

4(ac + bd)^{2 }- 4(a^{2} + b^{2}) (c^{2} + d^{2}) < 0

a^{2}c^{2} + b^{2}d^{2} + 2abcd - (a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2}) < 0

a^{2}c^{2} + b^{2}d^{2} + 2abcd - a^{2}c^{2} - a^{2}d^{2} - b^{2}c^{2} - b^{2}d^{2} < 0

2abcd - a^{2}d^{2} - b^{2}c^{2} < 0

-(ad - bc)^{2 }< 0

(ad - bc)^{2 }> 0

For this condition to be true ad ≠ bc

So, the correct option is (d).

### Solution 17

### Solution 18

It is given that x = 1 is root of equation ax^{2} + 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 x^{2} + x + b = 0

Hence, (1)^{2 }+ (1) + b = 0

b = -2 ......(2)

from (1) & (2)

ab = (-1) (-2)

ab = 2

So, the correct option is (b).

### Solution 19

### Solution 20

### Solution 21

### Solution 22

### Solution 23

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

so (2)^{2 }+ a(2) + 12 = 0

4 + 2a + 12 = 0

a = -8 .....(1)

Given x^{2 }+ ax + q = 0 has equal roots so

a^{2 }- 4q = 0 ......(2)

from (1) & (2)

(-8)^{2 }- 4q = 0

4q = 64

q = 16

So, the correct option is (d).

### Solution 24

### Solution 25

If a and b are roots of the equation x^{2 }+ 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^{2 }- 4 = 0

So, the correct option is (b).

### Solution 27

### Solution 28

### Solution 29

x^{2 }+ ax + 3 = 0

product of roots = 3

One root is 1. Hence other root is 3.

So, the correct option is (a).

## Quadratic Equations Exercise 4.85

### Solution 30

### Solution 31

### Solution 32

Any quadratic equation, ax^{2 }+ bx + c = 0 has real and equal roots if b^{2 }- 4ac = 0

For the question, equation is 16x^{2 }+ 4kx + 9 = 0,

(4k)^{2 }- 4 × 16 × 9 = 0

16k^{2 }- 36 × 16 = 0

k^{2} - 36 = 0

k^{2 }= 36

k = ± 6

So, the correct option is (c).