NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 4
Chapter Name : Quadratic Equations

Exercise 4.1

Q1 Check whether the following are quadratic equations :
2 2
( i) (x + 1) = 2(x − 3) (ii) x − 2x = (−2)(3 − x)

( iii) (x − 2)(x + 1) = (x − 1)(x + 3) (iv) (x − 3)(2x + 1) = x(x + 5)
2 2
(v)(2x − 1)(x − 3) = (x + 5)(x − 1) (vi) x + 3x + 1 = (x − 2)
3 2 3 2 3
(vii)(x + 2) = 2x (x − 1) (viii) x − 4x − x + 1 = (x − 2)

2 2 2
(i) (x + 1) = 2(x − 3) ⇒ x + 2x + 1 = 2x − 6 ⇒ x + 7 = 0

2
It is of the form ax + bx + c = 0 .

Hence, the given equation is a quadratic equation.
2 2 2
(ii) x − 2x = (−2)(3 − x) ⇒ x − 2x = −6 + 2x ⇒ x − 4x + 6 = 0

2
It is of the form ax + bx + c = 0 .

Hence, the given equation is a quadratic equation.
2 2
(iii) (x − 2)(x + 1) = (x − 1)(x + 3) ⇒ x − x − 2 = x + 2x − 3 ⇒ 3x − 1 = 0

2
It is not of the form ax + bx + c = 0

Hence, the given equation is not a quadratic equation.
2 2 2
(iv) (x − 3)(2x + 1) = x(x + 5) ⇒ 2x − 5x − 3 = x + 5x ⇒ x − 10x − 3 = 0

2
It is of the form ax + bx + c = 0 .

Hence, the given equation is a quadratic equation.
2 2 2
(v) (2x − 1)(x − 3) = (x + 5)(x − 1) ⇒ 2x − 7x + 3 = x + 4x − 5 ⇒ x − 11x + 8 = 0

2
of the form ax + bx + c = 0 .

Hence, the given equation is a quadratic equation.
2 2 2 2
(vi) x + 3x + 1 = (x − 2) ⇒ x + 3x + 1 = x + 4 − 4x ⇒ 7x − 3 = 0

2
It is not of the form ax + bx + c = 0

Hence, the given equation is not a quadratic equation.

3 2 3 2 3 3 2
(vii) (x + 2) = 2x (x − 1) ⇒ x + 8 + 6x + 12x = 2x − 2x ⇒ x − 14x − 6x − 8 = 0

2
not of the form ax + bx + c = 0 .

Hence, the given equation is not a quadratic equation.
3 3 3 3 2 3 2 2
(viii) x − 4x − x + 1 = (x − 2) ⇒ x − 4x − x + 1 = x − 8 − 6x + 12x ⇒ 2x − 13x + 9 = 0

2
It is of the form ax + bx + c = 0

Hence, the given equation is a quadratic equation.

Page : 73 , Block Name : Exercise 4.1

Q2 Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m . The length of the plot (in metres) is one more than twice its
2

breadth. We need to nd the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to nd the integers. (iii) Rohan’s
mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We
would like to nd Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would
have taken 3 hours more to cover the same distance. We need to nd the speed of the train.

(i) Let the breadth of the plot be xm .

Hence, the length of the plot is (2x + 1)m .

Area of a rectangle = Length × Breadth

∴ 528 = x(2x + 1)

2
⇒ 2x + x − 528 = 0

(ii) Let the consecutive integers be x and x + 1

It is given that their product is 306 .

2
∴ x(x + 1) = 306 ⇒ x + x − 306 = 0

(iii) Let Rohan's age be x .

Hence, his mother's age = x + 26

3 years hence,

Rohan's age = x + 3

Mother's age = x + 26 + 3 = x + 29

It is given that the product of their ages after 3 years is 360 .

∴ (x + 3)(x + 29) = 360

2
⇒ x + 32x − 273 = 0

(iv) Let the speed of train be xkm/h .

480
Time taken to travel 480km = hrs
x

In second condition, let the speed of train = (x − 8)km/h

It is also given that the train will take 3 hours to cover the same

distance.

480
Therefore, time taken to travel 480km = ( + 3)
x
hrs

Speed × Time = Distance

480
(x − 8) ( + 3) = 480
x

3840
⇒ 480 + 3x − − 24 = 480
x

3840
⇒ 3x − = 24
x

2
⇒ 3x − 24x + 3840 = 0

2
⇒ x − 8x + 1280 = 0

Page : 73 , Block Name : Exercise 4.1

Exercise 4.2

Q1 Find the roots of the following quadratic equations by factorisation:
2 2
(i) x − 3x − 10 = 0 (ii) 2x + x − 6 = 0

2
(iii) √2x + 7x + 5√2 = 0

2 2 1
(v) 100x − 20x + 1 = 0 (iv) 2x − x + = 0
8

2
(i)x − 3x − 10
2
= x − 5x + 2x − 10

= x(x − 5) + 2(x − 5)

= (x − 5)(x + 2)

Roots of this equation are the values for which (x − 5)(x + 2) = 0

∴ x − 5 = 0 or x + 2 = 0

i.e., x = 5 or x = −2

2
( ii) 2x + x − 6
2
= 2x + 4x − 3x − 6

= 2x(x + 2) − 3(x + 2)

= (x + 2)(2x − 3)

Roots of this equation are the values for which (x + 2)(2x − 3) = 0

∴ x + 2 = 0 or 2x − 3 = 0

3
i.e., x = −2 or x =
2

2
(iii) √2x + 7x + 5√2
2
= √2x + 5x + 5x + 5√2

= x(√2x + 5) + √2(√2x + 5)

= (√2x + 5)(x + √2)

Roots of this equation are the values for which (√2x + 5)(x + √2) = 0

∴ √2x + 5 = 0 or x + √2 = 0

−5
i.e., x = or x = −√2
√2

(iv)
1
2
2x − x +
8

1
2
= (16x − 8x + 1)
8

1
2
= (16x − 4x − 4x + 1)
8

1
= (4x(4x − 1) − 1(4x − 1))
8

1
2
= (4x − 1)
8

2
Roots of this equation are the values for which (4x − 1) = 0

Therefore, (4x − 1) = 0 or (4x − 1) = 0

1 1
i.e., x = or x =
4 4

2
(v)100x − 20x + 1

2
= 100x − 10x − 10x + 1

= 10x(10x − 1) − 1(10x − 1)

2
= (10x − 1)
2
Roots of this equation are the values for which (10x − 1) = 0

Therefore, (10x − 1) = 0 or (10x − 1) = 0

1 1
i.e., x = or x =
10 10

Page : 76 , Block Name : Exercise 4.2

Q2 Solve the problems:
Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the
number of marbles they now have is 124. We would like to nd out how many marbles they had to start
with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in
rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost
of production was Rs. 750. We would like to nd out the number of toys produced on that day.

(i) Let the number of John's marbles be x .

The refore, number of Jivanti's marble = 45 − x

After losing 5 marbles,

Number of John's marbles = x − 5

Number of Jivanti's marbles = 45 − x − 5 = 40 − x

It is given that the product of their marbles is 124.

∴ (x − 5)(40 − x) = 124

2
⇒ x − 45x + 324 = 0

2
⇒ x − 36x − 9x + 324 = 0

⇒ x(x − 36) − 9(x − 36) = 0

⇒ (x − 36)(x − 9) = 0

Either x − 36 = 0 or x − 9 = 0

i.e., x = 36 or x = 9

If the number of John's marbles = 36 ,

Then, number of Jivanti's marbles = 45 − 36 = 9

If number of John's marbles = 9,

Then, number of Jivanti's marbles = 45 − 9 = 36

(ii) Let the number of toys produced be x .

∴ cost of production of each toy = Rs(55 − x)

It is given that, total production of the toys = Rs750

∴ x(55 − x) = 750

2
⇒ x − 55x + 750 = 0

2
⇒ x − 25x − 30x + 750 = 0

⇒ x(x − 25) − 30(x − 25) = 0

⇒ (x − 25)(x − 30) = 0

Either x − 25 = 0 or x − 30 = 0

i.e., x = 25 or x = 30

Hence, the number of toys will be either 25 or 30 .

Page : 76 , Block Name : Exercise 4.2

Q3 Find two numbers whose sum is 27 and product is 182.

Let the first number be x and the second number is 27 − x .

Therefore, their product = x(27 − x)
It is given that the product of these numbers is 182 .

Therefore, x(27 − x) = 182
2
⇒ x − 27x + 182 = 0

2
⇒ x − 13x − 14x + 182 = 0

⇒ x(x − 13) − 14(x − 13) = 0

⇒ (x − 13)(x − 14) = 0

Either x − 13 = 0 or x − 14 = 0

l.e., x = 13 or x = 14

If first number = 13 , then

other number = 27 − 13 = 14

If first number = 14, then

If first number = 27 − 14 = 13

Therefore, the numbers are 13 and 14.

Page : 76 , Block Name : Exercise 4.2

Q4 Find two consecutive positive integers, sum of whose squares is 365.

Let the consecutive positive integers be x and x + 1.

2 2
Given that x + (x + 1) = 365

2 2
⇒ x + x + 1 + 2x = 365

2
⇒ 2x + 2x − 364 = 0

2
⇒ x + x − 182 = 0

2
⇒ x + 14x − 13x − 182 = 0

⇒ x(x + 14)(x − 13) = 0

⇒ (x + 14)(x − 13) = 0

Either x + 14 = 0 or x − 13 = 0, l.e., x = −14 or x = 13

since the integers are positive, x can only be 13.

∴ x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

Page : 76 , Block Name : Exercise 4.2

Q5 The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, nd the other two
sides.

Let the base of the right triangle be xcm.

Its altitude = (x − 7)cm

From pythagoras theorem,

2 2 2
Base + Altitude = Hypotenuse

2 2 2
∴ x + (x − 7) = 13

2 2
⇒ x + x + 49 − 14x = 169

2
⇒ 2x − 14x − 120 = 0

2
⇒ x − 7x − 120 = 0

2
⇒ x − 7x − 120 = 0

2
⇒ x − 7x − 60 = 0

2
⇒ x − 12x − 5x − 60 = 0

⇒ x(x − 12) + 5(x − 12) = 0

⇒ (x − 12)(x + 5) = 0

Either x − 12 = 0 or x + 5 = 0, i.e., x = 12 or x = −5

since sides are positive, x can only be 12 .

Therefore, the base of the given triangle is 12cm and the altitude of

this triangle will be (12 − 7)cm = 5cm.

Page : 76 , Block Name : Exercise 4.2

Q6 A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular
day that the cost of production of each article (in rupees) was 3 more than twice the number of articles
produced on that day. If the total cost of production on that day was Rs 90, nd the number of articles
produced and the cost of each article.

Let the number of articles produced be x .

Therefore, cost of production of each article = Rs (2x + 3)

It is given that the total production is Rs 90.

∴ x(2x + 3) = 90

2
⇒ 2x + 3x − 90 = 0

2
⇒ 2x + 15x − 12x − 90 = 0

⇒ x(2x + 15) − 6(2x + 15) = 0

⇒ (2x + 15)(x − 6) = 0

−15
Either 2x + 15 = 0 or x − 6 = 0, i.e., x = or x = 6
2

As the number of articles produced can only be a positive integer,

therefore, x can only be 6 .

Hence, number of articles produced = 6

cost of each article = 2 × 6 + 3 = Rs 15

Page : 76 , Block Name : Exercise 4.2

Exercise 4.3

Q1 Find the roots of the following quadratic equations, if they exist, by the method of completing the
square:
2
(ii) 2x + x − 4 = 0
2 2
(i) 2x − 7x + 3 = 0 (iii) 4x + 4√3x + 3 = 0
2
(iv) 2x + x + 4 = 0

Answer. (i)
2
2x − 7x + 3 = 0

2
⇒ 2x − 7x = −3

On dividing both sides of the equation by 2, we obtain

2 7 3
⇒ x − x = −
2 2

2 7 3
⇒ x − 2 × x × = −
4 2

2
7
On adding ( ) to both sides of equation, we obtain
4

2 2
2 7 7 7 3
⇒ (x) − 2 × x × + ( ) = ( ) −
4 4 4 2

2
7 49 3
⇒ (x − ) = −
4 16 2

2
7 25
⇒ (x − ) =
4 16

7 5
⇒ (x − ) = ±
4 4

7 5
⇒ x = ±
4 4

7 5 7 5
⇒ x = + or x = −
4 4 4 4

12 2
⇒ x = or x =
4 4

1
⇒ x = 3 or
2
2
(ii) 2x + x − 4 = 0

2
On comparing this equation with ax + bx + c = 0, we obtain

a = 2, b = 1, c = −4

By using quadratic formula, we obtain

−b±√b2 −4ac
x =
2a

−1±√1+32
⇒ x =
4

−1±√33
⇒ x =
4

−1+√33 −1−√33
∴ x = or
4 4

2
(iii) 4x + 4√3x + 3 = 0

2
On comparing this equation with ax + bx + c = 0, we obtain

q = 4, b = 4√3, c = 3

By using quadratic formula, we obtain

−b±√b2 −4ac
x =
2a

−4√3±√48−48
⇒ x =
8

−4√3±0
⇒ x =
8

−√3 −√3
∴ x = or
2 2
2
(iv) 2x + x + 4 = 0

2
On comparing this equation with ax + bx + c = 0, we obtain

a = 2, b = 1, c = 4

By using quadratic formula, we obtain

−b±√b2 −4ac
x =
2a

−1±√1−32
⇒ x =
4

−1±√−31
⇒ x =
4

However, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.

Page : 87 , Block Name : Exercise 4.3

Q2 Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

2
(i) 2x − 7x + 3 = 0

2
On comparing this equation with ax + bx + c = 0, we obtain

a = 2, b = −7. c = 3

By using quadratic formula, we obtain

−b±√b2 −4ac
x =
2a

7±√49−24
⇒ x =
4

7±√25
⇒ x =
4

7±5
⇒ x =
4

7+5 7−5
⇒ x = or
4 4

12 2
⇒ x = or
4 4

1
∴ x = 3 or
2

2
(ii) 2x + x − 4 = 0

2
On comparing thisequation with ax + bx + c = 0, we obtain

a = 2, b = 1, c = −4

By using quadratic formula, we obtain

−b±√b2 −4ac
x =
2a

−1±√1+32
⇒ x =
4

−1±√33
⇒ x =
4

−1+√33 −1−√33
∴ x = or
4 4

2
(iii) 4x + 4√3x + 3 = 0

2
On comparing this equation with ax + bx + c = 0, we obtain

a = 4, b = 4√3, c = 3

By using quadratic formula, we obtain

2
−b±√b −4ac
x =
2a

−4√3±√48−48
⇒ x =
8

−4√3±0
⇒ x =
8

−√3 −√3
∴ x = or
2 2
2
(iv) 2x + x + 4 = 0

2
On comparing this equation with ax + bx + c = 0, we obtain

a = 2, b = 1, c = 4

By using quadratic formula, we obtain

2
−b±√b −4ac
x =
2a

−1±√1−32
⇒ x =
4

−1±√−31
⇒ x =
4

However, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.

Page : 87 , Block Name : Exercise 4.3

Q3 Find the roots of the following equations:
(i) x − = 3, x ≠ 01

x

(ii) 1

x+4
−
x−7
1
=
11

30
, x ≠ −4, 7

1 2
(i) x − = 3 ⇒ x − 3x − 1 = 0
x

2
On comparing this equation with ax + bx + c = 0, we obtain

a = 1, b = −3, c = −1

By using quadratic formula, we obtain

2
−b±√b −4ac
x =
2a

3±√9+4
⇒ x =
2

3±√13
⇒ x =
2

3+√13 3−√13
Therefore, x = or
2 2

1 1 11
(ii) − =
x + 4 x − 7 30

x − 7 − x − 4 11
⇒ =
(x + 4)(x − 7) 30

−11 11
⇒ =
(x + 4)(x − 7) 30

⇒ (x + 4)(x − 7) = −30

2
⇒ x − 3x − 28 = −30
2
⇒ x − 3x + 2 = 0

2
⇒ x − 2x − x + 2 = 0

⇒ x(x − 2) − 1(x − 2) = 0

⇒ (x − 2)(x − 1) = 0

⇒ x = 1 or 2

Page : 88 , Block Name : Exercise 4.3

Q4 The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1

3
Find his
present age.

Let the present age of Rehman be x years.

Three years ago, his age was (x − 3) years.

Five years hence, his age will be (x + 5) years.

It is given that the sum of the reciprocals of Rehman's ages 3 years

1
ago and 5 years from now is
3

1 1 1
∴ + =
x−3 x+5 3

x+5+x−3 1
=
(x−3)(x+5) 3

2x+2 1
=
(x−3)(x+5) 3

⇒ 3(2x + 2) = (x − 3)(x + 5)
2
⇒ 6x + 6 = x + 2x − 15

2
⇒ x − 4x − 21 = 0

2
⇒ x − 7x + 3x − 21 = 0

⇒ x(x − 7) + 3(x − 7) = 0

⇒ (x − 7)(x + 3) = 0

⇒ x = 7, −3

However, age cannot be negative.

Therefore, Rehman's present age is 7 years.

Page : 88 , Block Name : Exercise 4.3

Q5 In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more
in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her
marks in the two subjects.

Let the marks in Maths be x .

Then, the marks in English will be 30 − x .

According to the given question,

(x + 2)(30 − x − 3) = 210

(x + 2)(27 − x) = 210
2
⇒ −x + 25x + 54 = 210

2
⇒ x − 25x + 156 = 0

2
⇒ x − 12x − 13x + 156 = 0

⇒ x(x − 12) − 13(x − 12) = 0

⇒ (x − 12)(x − 13) = 0

⇒ x = 12, 13

If the marks in Maths are 12, then marks in English will be 30 — 12 18 If the marks in Maths are 13, then
marks in English will be 30 — 13 17

Page : 88 , Block Name : Exercise 4.3

Q6 The diagonal of a rectangular eld is 60 metres more than the shorter side. If the longer side is 30
metres more than the shorter side, nd the sides of the eld.

Let the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30)m

2 2
Diagonal of the rectangle = √x + (x + 30)

It is given that the diagonal of the rectangle is 60m more than the shorter side.

2 2
∴ √x + (x + 30) = x + 60

2 2 2
⇒ x + (x + 30) = (x + 60)

2 2 2
⇒ x + x + 900 + 60x = x + 3600 + 120x

2
⇒ x − 60x − 2700 = 0

2
⇒ x − 90x + 30x − 2700 = 0

⇒ (x − 90)(x + 30) = 0

⇒ x = 90, −30

However, side cannot be negative. Therefore, the length of the shorter side will be 90 m. Hence, length Of
the larger side will be (90 + 30) m = 120 m

Page : 88 , Block Name : Exercise 4.3

Q7 The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger
number. Find the two numbers.

Answer. Let the larger and smaller number be x and y respectively. According to the given question,

2 2 2
x − y = 180 and y = 8x

2
⇒ x − 8x = 180

2
⇒ x − 8x − 180 = 0

2
⇒ x − 18x + 10x − 180 = 0

⇒ x(x − 18) + 10(x − 18) = 0

⇒ (x − 18)(x + 10) = 0

⇒ x = 18, −10

However,the larger number can not be negative as 8 times of the larger number will be negative and hence,
the square of the smaller number will be negative which is not possible. Therefore, the larger number will
be 18 only.
x = 18

2
∴ y = 8x = 8 × 18 = 144

⇒ y = ±√144 = ±12

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and − 12 .

Page : 88 , Block Name : Exercise 4.3

Q8 A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1
hour less for the same journey. Find the speed of the train.

Let the speed of the train be xkm/hr .

360
Time taken to cover 360km =
x hr

According to the given question,

360
(x + 5) ( − 1) = 360
x

360
⇒ (x + 5) ( − 1) = 360
x

1800
⇒ 360 − x + − 5 = 360
x

2
⇒ x + 5x − 1800 = 0

2
⇒ x + 45x − 1800 = 0

⇒ x(x + 45)(x − 40) = 0

⇒ (x + 45)(x − 40) = 0

⇒ x = 40, −45

However, speed cannot be negative.Therefore, the speed Of train is 40 km/h.

Page : 88 , Block Name : Exercise 4.3

Q9 Two water taps together can ll a tank in 9$ hours. The tap of larger diameter takes 10 hours less
3
$
8

than the smaller one to ll the tank separately. Find the time in which each tap can separately ll the tank.

Answer. Let the time taken by the smaller pipe to ll the tank be x hr. Time taken by the larger pipe = (x—
10) hr
Part of tank lled by smaller pipe in 1 hour = 1

x

Part of tank lled by smaller pipe in 1 hour = x−10
1

3 75
It is given that the tank can be filled in 9 = hours by both the pipes
8 8

together. Therefore,

1 1 8
+ =
x x−10 75

x−10+x 8
=
x(x−10) 75

2x−10 8
⇒ =
x(x−10) 75

2
⇒ 75(2x − 10) = 8x − 80x

2
⇒ 150x − 750 = 8x − 80x

2
⇒ 8x − 230x + 750 = 0

2
⇒ 8x − 200x − 30x + 750 = 0

⇒ 8x(x − 25) − 30(x − 25) = 0

⇒ (x − 25)(8x − 30) = 0

30
i.e.. x = 25,
8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours. As in this case, the time taken by the larger
pipe will be negative, which is logically not possible. Therefore, time taken individually by the smaller pipe
and the larger pipe Will be 25 and 25 - 10 = 15 hours respectively.

Page : 88 , Block Name : Exercise 4.3

Q10 An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and
Bangalore (without taking into consideration the time they stop at intermediate stations). If the average
speed of the express train is 11km/h more than that of the passenger train, nd the average speed of the
two trains. 11. Sum of the areas of two squares is 468 m² . If the difference of their perimeters is 24 m, nd
the sides of the two squares.

Answer. Let the average speed of passenger train be x km/h. Average speed of express train = (x + 11) km/h
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to
cover the same distance.
132 132
∴ − = 1
x x+11

x+11−x
⇒ 132 [ ] = 1
x(x+11)

132×11
⇒ = 1
x(x+11)

⇒ 132 × 11 = x(x + 11)

2
⇒ x + 11x − 1452 = 0
2
⇒ x + 44x − 33x − 1452 = 0

⇒ x(x + 44) − 33(x + 44) = 0

⇒ (x + 44)(x − 33) = 0

⇒ (x + 44)(x − 33) = 0

⇒ x = −44, 33

Speed cannot be negative.

Therefore, the speed of the passenger train will be 33km/h and thus,

the speed of the express train will be 33 + 11 = 44km/h .

Page : 88 , Block Name : Exercise 4.3

Q11 Sum of the areas of two squares is 468 m² . If the difference of their perimeters is 24 m, nd the sides
of the two squares.

Answer. Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y
respectively and their areas will be x and y respectively.2 2

It is given that
4x − 4y = 24

x − y = 6

x = y + 6

2 2
Also, x + y = 468

2 2
⇒ (6 + y) + y = 468

2 2
⇒ 36 + y + 12y + y = 468
2
⇒ 2y + 12y − 432 = 0

2
⇒ y + 6y − 216 = 0

2
⇒ y + 18y − 12y − 216 = 0

⇒ y(y + 18) − 12(y + 18) = 0

⇒ (y + 18)(y − 12) = 0

⇒ y = −18 or 12

However, side Of a square cannot be negative, Hence, the sides of the squares are 12 m and (12 + 6) m = 18
m.

Page : 88 , Block Name : Exercise 4.3

Exercise 4.4

Q1 Find the nature of the roots of the following quadratic equations. If the real roots exist, nd them:
2
(I) 2x − 3x + 5 = 0

2
(II) 3x − 4√3x + 4 = 0

2
(III) 2x − 6x + 3 = 0

Answer. We know that for a quadratic equation ax 2
+ bx + c = 0 , discriminant is b² - 4ac.
2
(A) If b − 4ac > 0 → two distinct real roots

2
(B) If b − 4ac = 0 → two equal real roots

2
(C) If b − 4ac < 0 → no real roots

2
(I) 2x − 3x + 5 = 0

2
Comparing this equation with ax + bx + c = 0, we obtain

a = 2, b = −3, c = 5

2 2
Discriminant = b − 4ac = (−3) − 4(2)(5) = 9 − 40

= −31

2
As b − 4ac < 0

Therefore, no real root is possible for the given equation.

2
Comparing this equation with ax + bx + c = 0, we obtain

2
Comparing this equation with ax + bx + c = 0, we obtain

a = 3, b = −4√3, c = 4
2 2
Discriminant = b − 4ac = (−4√3) − 4(3)(4)

= 48 − 48 = 0

2
As b − 4ac = 0

Therefore, real roots exist for the given equation and they are equal to

each other.

−b −b
And the roots will be and
2a 2a

−b −(−4√3) 4√3 2√3 2
= = = =
2a 2×3 6 3 √3

2
Therefore, the roots are √3 and
√3

2
(III) 2x − 6x + 3 = 0

2
Comparing this equation with ax + bx + c = 0, we obtain

a = 2, b = −6, c = 3

2 2
Discriminant = b − 4ac = (−6) − 4(2)(3)

= 36 − 24 = 12

2
As b − 4ac > 0

Therefore, distinct real roots exist for this equation as follows.

−b±√b2 −4ac
x =
2a

2
−(−6)±√(−6) −4(2)(3)
=
2(2)

6±√12 6±2√3
= =
4 4

3±√3
=
2

3+√3 3−√3
Therefore, the roots are or
2 2

Page : 91 , Block Name : Exercise 4.4

Q2 Find the values of k for each of the following quadratic equations, so that they have two equal roots.
2
(I) 2x + kx + 3 = 0

(II) kx(x − 2) + 6 = 0

Answer. We know that if an equation ax 2
+ bx + c = 0 has two equal roots, its discriminant
2
(b − 4ac) will be 0.

2
(I) 2x + kx + 3 = 0

2
Comparing equation with ax + bx + c = 0, we obtain

a = 2, b = k, c = 3

2 2
Discriminant = b − 4ac = (k) − 4(2)(3)

2
= k − 24

For equal roots,

Discriminant = 0

2
k − 24 = 0

2
k = 24

k = ±√24 = ±2√6

(II) kx(x − 2) + 6 = 0

2
or kx − 2kx + 6 = 0

2
Comparing this equation with ax + bx + c = 0, we obtain

a = k, b = −2k, c = 6

2 2
Discriminant = b − 4ac = (−2k) − 4(k)(6)

2
= 4k − 24k

For equal roots,

2
b − 4ac = 0

2
4k − 24k = 0

4k(k − 6) = 0

Either 4k = 0 or k = 6 = 0

k = 0 or k = 6

However, if k = 0, then the equation will not have the terms x and x. 2

Therefore, if this equation has two equal roots, k should be 6 only.

Page : 91 , Block Name : Exercise 4.4

Q3 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800
m ? If so, nd its length and breadth.
2

Answer. Let the breadth of mango grove be l.Length of mango grove will be 21. Area of mango grove = (21)
(l)
2
= 2l

2
2l = 800

2 800
l = = 400
2

2
l − 400 = 0

2
Comparing this equation with al + bl + c = 0, we obtain

a = 1b = 0, c = 400

2 2
Discriminant = b − 4ac = (0) − 4 × (1) × (−400) = 1600

2
Here, b − 4ac > 0

Therefore, the equation will have real roots. And hence, the desired

rectangular mango grove can be designed.

l = ±20

However, length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40m

Page : 91 , Block Name : Exercise 4.4

Q4 Is the following situation possible? If so, determine their present ages. The sum of the ages of two
friends is 20 years. Four years ago, the product of their ages in years was 48.

Answer. Let the age of one friend be x years. Age of the other friend will be (20 - x) years. 4 years ago, age of
1st friend = (x - 4) years
nd
And, age of 2 friend = (20 − x − 4)

= (16 − x) years

Given that,

(x − 4)(16 − x) = 48

2
16x − 64 − x + 4x = 48

2
−x + 20x − 112 = 0

2
x − 20x + 112 = 0
2
Comparing this equation with ax + bx + c = 0, we obtain

a = 1, b = −20, c = 112

2 2
Discriminant = b − 4ac = (−20) − 4(1)(112)

= 400 − 448 = −48

2
As b − 4ac < 0

Therefore, no real root is possible for this equation and hence, this

situation is not possible.

Page : 91 , Block Name : Exercise 4.4

Q5 Is it possible to design a rectangular park of perimeter 80 and area 400m ? If so nd its length and
2

breadth.

Answer. Let the length and breadth of the park be I and b. Perimeter = 2 (l + b) = 80 l + b = 40 or, b = 40 - l
2
Area = l × b = I (40 − l) = 40l − l

2
40l − l = 400

P − 40l + 400 = 0

Comparing this equation with

af + bl + c = 0, we obtain

a = 1, b = −40, c = 400

2 2
Discriminate = b − 4ac = (−40) − 4(1)(400)

= 1600 − 1600 = 0

2
As b − 4ac = 0

Therefore, this equation has equal real roots. And hence, this situation

is possible.

Root of this equation,

b
l = −
2a

(−40) 40
l = − = = 20
2(1) 2

Therefore, length of park, I = 20m

And breadth of park, b = 40 − I = 40 − 20 = 20m

Page : 91 , Block Name : Exercise 4.4