Quadratic Equations

Class 10 > Math > NCERT > Quadratic Equations

NCERT Solutions for Chapter 4 Quadratic Equations Class 10 Maths

Book Solutions

1(i)

Check whether the following are quadratic equations:
(x + 1)2 = 2(x - 3)

Answer

We have:

(x + 1)2 = 2 (x − 3)
⇒ x2 + 2x + 1 = 2x − 6
⇒ x2 + 2x + 1 − 2x + 6 = 0
⇒ x2 + 7 = 0

Since x2 + 7 is a quadratic polynomial
∴ (x + 1)2 = 2(x − 3) is a quadratic equation.

Exercise 4.1 Page Number 73

1(ii)

x2 - 2x = (-2)(3 - x)

Answer

We have:

x2 − 2x = (− 2) (3 − x)
⇒ x2 − 2x = − 6 + 2x
⇒ x2 − 2x − 2x + 6 = 0
⇒ x2 − 4x + 6 = 0

Since x2 − 4x + 6 is a quadratic polynomial
∴ x2 − 2x = (−2) (3 − x) is a quadratic equation.

Exercise 4.1 Page Number 73

1(iii)

(x - 2)(x + 1) = (x - 1)(x + 3)

Answer

We have:

(x − 2) (x + 1) = (x − 1) (x + 3)
⇒ x2 − x − 2 = x2 + 2x − 3
⇒ x2 − x − 2 − x2 − 2x + 3 = 0
⇒ −3x + 1 = 0

Since −3x + 1 is a linear polynomial

∴ (x − 2) (x + 1) = (x − 1) (x + 3) is not quadratic equation.

Exercise 4.1 Page Number 73

1(iv)

(x - 3)(2x + 1) = x(x + 5)

Answer

We have:

(x − 3) (2x + 1) = x (x + 5)
⇒ 2x2 + x − 6x −3= x2 + 5x
⇒ 2x2 − 5x − 3 − x2 − 5x =0
⇒ x2 + 10x − 3= 0

Since x2 + 10x − 3 is a quadratic polynomial

∴ (x − 3) (2x + 1) = x(x + 5) is a quadratic equation.
Exercise 4.1 Page Number 73

1(v)

(2x - 1)(x - 3) = (x + 5)(x - 1)

Answer

We have:

(x − 3) (2x + 1) = x (x + 5)
⇒ 2x2 + x − 6x −3= x2 + 5x
⇒ 2x2 − 5x − 3 − x2 − 5x =0
⇒ x2 + 10x − 3= 0

Since x2 + 10x − 3 is a quadratic polynomial

∴ (x − 3) (2x + 1) = x(x + 5) is a quadratic equation.
Exercise 4.1 Page Number 73

1(vi)

x2 + 3x + 1 = (x - 2)2

Answer

We have:

x2 + 3x + 1 = (x − 2)2
⇒ x2 + 3x + 1 = x2 − 4x + 4
⇒ x2 + 3x + 1 − x2 + 4x −4= 0
⇒ 7x − 3 = 0

Since 7x − 3 is a linear polynomial

∴ x2 + 3x + 1 = (x − 2)2 is not a quadratic equation.

Exercise 4.1 Page Number 73

1(vii)

(x + 2)3 = 2x(x2 - 1)

Answer

We have:

(x + 2)3 =2x(x2 − 1)
⇒ x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 − 2x
⇒ x3 + 6x2 + 12x + 8 = 2x3 − 2x
⇒ x3 + 6x2 + 12x + 8 − 2x3 + 2x = 0
⇒− x3 + 6x2 + 14x + 8 = 0
Since −x3 + 6x2 + 14x + 8 is a polynomial of degree 3
∴ (x + 2)3 = 2x(x2 − 1) is not a quadratic equation.
Exercise 4.1 Page Number 73

1(viii)

x3 - 4x2 - x + 1 = (x - 2)3

Answer

We have:

(x + 2)3 =2x(x2 − 1)
⇒ x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 − 2x
⇒ x3 + 6x2 + 12x + 8 = 2x3 − 2x
⇒ x3 + 6x2 + 12x + 8 − 2x3 + 2x = 0
⇒− x3 + 6x2 + 14x + 8 = 0
Since −x3 + 6x2 + 14x + 8 is a polynomial of degree 3
∴ (x + 2)3 = 2x(x2 − 1) is not a quadratic equation.
Exercise 4.1 Page Number 73

2(i)

Represent the following situations in the form of quadratic equations.

(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Answer

Let the breadth = x metres
∵ Length = 2 (Breadth) + 1

∴ Length = (2x + 1) metres

Since Length × Breadth = Area

∴ (2x + 1) × x = 528
⇒ 2x2 + x = 528
⇒ 2x2 + x − 528 = 0
Thus, the required quadratic equation is 2x2 + x− 528 = 0

Exercise 4.1 Page Number 73

2(ii)

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Answer

Let the two consecutive numbers be x and (x + 1).
∵Product of the numbers = 306
∴ x (x + 1) = 306
⇒ x2 + x = 306
⇒ x2 + x − 306 = 0

Thus, the required equdratic equation is x2 + x − 306 = 0
Exercise 4.1 Page Number 74

2(iii)

(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 find Rohan's present age.

Answer

Let the present age = x
∴ Mother’s age = (x + 26) years

After 3 years

His age = (x + 3) years
Mother’s age = [(x + 26) + 3] years
=(x + 29) years

According to the condition,

[Product of their ages after 3 years] = 360

⇒ (x + 3) × (x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 29x + 3x + 87 − 360 = 0
⇒ x2 + 32x − 273 = 0

Thus, the required quadratic equation is

x2 + 32x − 273 = 0

Exercise 4.1 Page Number 74

2(iv)

(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 find the speed of the train.

Answer

In first case,

Let the speed of the train = u km/hr
Distance covered = 480 km
Time taken = Distance ÷ Speed

Thus, the required quadratic equation is u2 − 8u – 1280 = 0

Exercise 4.1 Page Number 74

1(i)

Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0

Answer

We have:
x2 − 3x − 10 = 0
⇒ x2 − 5x + 2x − 10 = 0
⇒ x (x − 5) + 2 (x − 5) = 0
⇒ (x − 5) (x + 2) = 0 Either x − 5=0
⇒ x = 5 or x + 2 = 0
⇒ x = − 2

Thus, the required roots are x = 5 and x = −2.

Exercise 4.2 Page Number 76

1(ii)

(ii) 2x2 + x – 6 = 0

Answer

We have:

2x2 + x − 6 = 0
⇒ 2x2 + 4x − 3x − 6 =0
⇒ 2x(x + 2) − 3 (x + 2) = 0
⇒ (x + 2) (2x − 3) = 0
Either x + 2 = 0 ⇒ x = −2

or 2x − 3= 0 ⇒ x = 3/2

Thus, the required roots are x = −2 and x = 3/2
Exercise 4.2 Page Number 76

1(iii)

(iii) √2 x2 + 7x + 5√2 = 0

Answer

We have:

Exercise 4.2 Page Number 76

1(iv)

(iv) 2x2 – x + 1/8 = 0

Answer

We have:

Thus, the required roots are = x = 1/4 and x = 1/4

Exercise 4.2 Page Number 76

1(v)

(v) 100x2 – 20x + 1 = 0

Answer

We have:

100x2 − 20x + 1 = 0
⇒ 100x2 − 10x − 10x + 1 = 0
⇒ 10x (10x − 1) − 1 (10x − 1) = 0
⇒ (10x − 1) (10x − 1) = 0
⇒ (10x − 1) = 0 and (10x − 1) = 0

⇒ x = 1/10 and x = 1/10

Thus, the required roots are x = 1/10 and x = 1/10.
Exercise 4.2 Page Number 76

2

(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. Find 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. Find out the number of toys produced on that day.

Answer

(i) Let the number of John's marbles be x.
Therefore, 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
⇒ x2 – 45x + 324 = 0
⇒ x2 – 36x - 9x + 324 = 0
⇒ x(x - 36) -9(x - 36) = 0
⇒ (x - 36)(x - 9) = 0   [∵ (-9) × (-36) = 324 and (-9)+(-36)=-4]
Either x - 36 = 0 or x - 9 = 0
⇒ 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 = Rs 750
∴ x(55 - x) = 750
⇒ x2 – 55x + 750 = 0
⇒ x2 – 25x - 30x + 750 = 0
⇒ x(x - 25) -30(x - 25) = 0
⇒ (x - 25)(x - 30) = 0   [∵ (-30)×(-25)= 750 and (-30)+(-25)= -55]
Either, x -25 = 0 or x - 30 = 0
⇒ x = 25 or x = 30
Hence, the number of toys will be either 25 or 30.
Exercise 4.2 Page Number 76

3

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

Answer

Here, sum of the numbers is 27.
Let one of the numbers be x.
∴ Other number = 27 − x
According to the condition,
Product of the numbers = 182

⇒ x (27 − x) = 182
⇒ 27x − x2 = 182
⇒− x2 + 27x − 182 = 0
⇒ x2 − 27x + 182 = 0
⇒ x2 – 13x - 14x + 182 = 0
⇒ x(x - 13) -14(x - 13) = 0
⇒ (x - 13)(x -14) = 0   [∵ −27 = (−13)+(−14) and (-13)×(-14)= 182]
Either x = -13 = 0 or x - 14 = 0
⇒ x = 13 or x = 14
If first number = 13, then
Other number = 27 - 13 = 14
If first number = 14, then
Other number = 27 - 14 = 13
Therefore, the numbers are 13 and 14.
Exercise 4.2 Page Number 76

4

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

Answer

Let the two consecutive positive integers be x and (x + 1).

Since the sum of the squares of the numbers = 365 

∴ x2 + (x + 1)2 = 365
⇒ x2 + [x2 + 2x + 1] = 365
⇒ x2 + x2 + 2x + 1 = 365
⇒ 2x2 + 2x + 1 − 365 = 0
⇒ 2x2 + 2x − 364 = 0
⇒ x2 + x − 182 = 0

⇒ x2 +14x −13x −182 = 0 

⇒ x(x+14) − 13(x+14)=0  [∵ +14 −13= 1and 14×(-13) = −182]

⇒ (x+14)(x−13) = 0

Either x + 14 = 0 ⇒ x = − 14
or x − 13 = 0 ⇒ x = 13
Since x has to be a positive integer
∴ x = 13
⇒ x + 1 = 13 + 1 = 14

Thus, the required consecutive positive integers are 13 and 14.
Exercise 4.2 Page Number 76

5

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

Answer

Let the base of the right triangle be x cm.
Its altitude = (x - 7) cm
From Pythagoras theorem, we have


Squaring both sides, we get
∴ x2 + (x - 7)2 = 169
⇒ x2 + x2 + 49 - 14x = 169
⇒ 2x2 - 14x - 120 = 0
⇒ x2 - 7x - 60 = 0
⇒ x2 - 12x + 5x - 60 = 0
⇒ x(x - 12) + 5(x - 12) = 0
⇒ (x - 12)(x + 5) = 0
Either x - 12 = 0 or x + 5 = 0,
⇒ x = 12 or x = - 5
Since sides are positive, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 - 7) cm = 5 cm.
Exercise 4.2 Page Number 76

6

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, find the number of articles produced and the cost of each article.

Answer

Let the number of articles produced in a day = x
∴ Cost of production of each article = ₹ (2x + 3)
According to the condition,

Total cost = 90
⇒ x × (2x + 3) = 90
⇒ 2x2 + 3x = 90
⇒ 2x2 + 3x − 90 = 0
⇒ 2x2 − 12x + 15x − 90 = 0
⇒ 2x (x − 6) + 15 (x − 6) = 0
⇒ (x − 6) (2x + 15) = 0

Either x − 6 = 0 ⇒ x = 6

or 2x + 15 = 0 ⇒ x = -15/2

But the number of articles cannot be negative.

x = -15/2 is not required

⇒ x = 6

∴ Cost of each article = ₹ (2 × 6 + 3) = ₹ 15
Thus, the required number of articles produced is 6 and the cost of each article is ₹15.
Exercise 4.2 Page Number 76

1(i)

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x2 – 7x +3 = 0

Answer

2x2−7x + 3 = 0

Dividing throughout by the co-efficient of x2, we get


Thus required roots are, x = 3 and x = 1/2

Exercise 4.3 Page Number 87

1(ii)

(ii) 2x2 + x – 4 = 0

Answer

We have:
2x2 + x − 4 = 0
Dividing throughout by 2,


Exercise 4.3 Page Number 87

1(iii)

(iii) 4x2 + 4√3x + 3 = 0

Answer

Exercise 4.3 Page Number 87

1(iv)

(iv) 2x2 + x + 4 = 0

Answer

2x2 + x + 4 = 0
Dividing throughout by 2, we have:

So, no real roots exist.

There is no real value of x satisfying the given equation.

Exercise 4.3 Page Number 87

2

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

Answer

(i) 2x2−7x + 3 = 0

Comparing the given equation with ax2 + bx + c = 0, we have:

a = 2
b = − 7
c = 3

∴ b2 − 4ac =(−7)2 − 4(2)(3)
⇒ 49 − 24 = 25 ≥ 0

Since b2 − 4ac > 0

∴ The given equation has real roots.

The roots are given by,

 

(ii) 2x2 + x−4 = 0

Comparing the given equation with ax2 + bx + c = 0, we have:

a = 2
b = 1 c = − 4
∴ b2 − 4ac = (1)2 − 4 (2) (− 4)

= 1 + 32
= 33 > 0

Since b2 − 4ac > 0
∴ The given equation has equal roots.

The roots are given by


(iii) 4x2 + 4√3x + 3 = 0

Comparing the given equation with ax2 + bx + c = 0, we have:

a = 4
b = 43
c = 3

 

(iv) 2x2 + x + 4 = 0

Comparing the given equation with ax2 + bx + c = 0, we have:

a = 2
b = 1
c = 4

∴ b2 − 4ac = (1)2 − 4(2)(4)
= 1 − 32 = − 31 < 0

Since b2 − 4ac is less than 0, therefore the given equation does not have real roots.

Exercise 4.3 Page Number 87

3(i)

Find the roots of the following equations:
(i) x-1/x = 3, x ≠ 0

Answer

x - 1/x = 3, x ≠ 0
Here, we have:

x−1/x = 3

⇒ x2 − 1= 3 x

⇒ x2 − 3x − 1 = 0   ...(1)

Comparing (1) with ax2 + bx + c = 0, we have:

a = 1
b = − 3
c = − 1

∴ b2 − 4ac =(− 3)2 − 4 (1) (− 1)
= 9 + 4 = 13 > 0

Exercise 4.3 Page Number 88

3(ii)

(ii) 1/x+4 - 1/x-7 = 11/30, x = -4, 7

Answer

⇒ − 11 × 30 = 11 (x2 − 3x − 28)
⇒ − 30 = x2 − 3x − 28
⇒ x2 − 3x − 28 + 30 = 0
⇒ x2 − 3x + 2 = 0

Comparing with ax2 + bx + c = 0, we have:

a =1
b = − 3
c = 2

∴ b2 − 4ac =(− 3)2 − 4(1)(2)
= 9 − 8 = 1 > 0

∴ The quadratic equation has real roots, which are given by:

Thus, the roots of the given equation are:

x = 2 and x = 1.
Exercise 4.3 Page Number 88

4

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.

Answer

Let the present age of Rehman = x
∴ 3 years ago Rehman’s age = (x − 3) years  
5  years later Rehman’s age = (x + 5) years

Now according to the condition,

⇒ 3 [x + 5 + x − 3] = (x − 3) (x + 5)
⇒ 3 [2x + 2] = x2 + 2x − 15
⇒ 6x + 6 = x2 + 2x − 15
⇒ x2 + 2x − 6x − 15 − 6=0
⇒ x2 − 4x − 21 = 0  ...(1)

Now, comparing (1) with ax2 + bx + c = 0, we have:

a =1
b = − 4
c = − 21

∴ b2 − 4ac =(− 4)2 − 4 (1)(−21)
= 16 + 84
= 100

Since age cannot be negative,

∴ x ≠− 3 ⇒ x = 7

So, the present age of Rehman = 7 years.

Exercise 4.3 Page Number 88

5

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.

Answer

Let, Shefali’s marks in Mathematics = x
∴ Marks in English = (30 − x) [∵ Sum of their marks in Eng. and Maths = 30]

Now, according to the condition,

(x + 2) × [(30 − x) − 3] = 210
⇒ (x + 2) × (30 − x − 3) = 210
⇒ (x + 2) (−x + 27) = 210
⇒ −x2 + 25x + 54 = 210
⇒ −x2 + 25x + 54 − 210 = 0
⇒ −x2 + 25x − 156 = 0
⇒ x2 − 25x + 156 = 0   ...(1)

Now, comparing (1) with ax2 + bx + c = 0

a = 1
b = −25
c = 156

∴ b2 − 4ac =(−25)2 − 4(1) (156)
= 625 − 624 = 1

When x = 13, then 30 − 13 = 17
When x = 12, then 30 − 12 = 18
Thus, marks in Maths = 13, marks in English = 17
marks in Maths = 12, marks in English = 18

Exercise 4.3 Page Number 88

6

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Answer

Let the shorter side (i.e., breadth) = x metres.
∴ The longer side (length) = (x + 30) metres.

In a rectangle,


⇒ (x + 60)2 = 2x2 + 60x + 900
⇒ x2 + 120x + 3600 = 2x2 + 60x + 900
⇒ 2x2 − x2 + 60x − 120x + 900 − 3600 = 0
⇒ x2 − 60x − 2700 = 0   ...(1)

Comparing (1) with ax2 + bx + c = 0

∴ a = 1
b = −60
c = −2700
∴ b2 − 4ac =(− 60)2 − 4 (1) (− 2700)
⇒ b2 − 4ac = 3600 + 10800
⇒ b2 − 4ac = 14400


Since breadth cannot be negative,

∴ x ≠−30 ⇒ x = 90
∴ x + 30 = 90 + 30 = 120
Thus, the shorter side = 90 m
The longer side = 120 m.

Exercise 4.3 Page Number 88

7

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 question,
x2 - y2 = 180 and y2 = 8x
⇒ x2 - 8x = 180
⇒ x2 - 8x - 180 = 0
⇒ x2 - 18x + 10x - 180 = 0
⇒ x(x - 18) +10(x - 18) = 0
⇒ (x - 18)(x + 10) = 0
⇒ x = 18, -10
However, the larger number cannot 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
∴ y2 = 8x = 8 × 18 = 144
⇒ y = ±√44 = ±12
∴ Smaller number = ±12
Therefore, the numbers are 18 and 12 or 18 and - 12
Exercise 4.3 Page Number 88

8

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.

Answer

Let the speed of the train be x km/hr.
Time taken to cover 360 km = 360/x hr.According to the question,
⇒ (x + 5)(360-1/x) = 360
⇒ 360 - x + 1800-5/x = 360
⇒ x2 + 5x + 10x - 1800 = 0
⇒ x(x + 45) -40(x + 45) = 0
⇒ (x + 45)(x - 40) = 0
⇒ x = 40, -45
However, speed cannot be negative.
Therefore, the speed of train is 40 km/h.
Exercise 4.3 Page Number 88

9

Two water taps together can fill a tank in hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer

Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x - 10) hr
Part of tank filled by smaller pipe in 1 hour = 1/x

Part of tank filled by larger pipe in 1 hour = 1/x - 10
It is given that the tank can be filled in  = 75/8 hours by both the pipes together. Therefore,
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
⇒ 2x-10/x(x-10) = 8/75
⇒ 75(2x - 10) = 8x2 - 80x
⇒ 150x - 750 = 8x2 - 80x
⇒ 8x2 - 230x +750 = 0
⇒ 8x2 - 200x - 30x + 750 = 0
⇒ 8x(x - 25) -30(x - 25) = 0
⇒ (x - 25)(8x -30) = 0
⇒ x = 25, 30/8
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.
Exercise 4.3 Page Number 88

10

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 speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

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 × 11 = x(x + 11)
⇒ x2 + 11x - 1452 = 0
⇒ x2 + 44x -33x -1452 = 0
⇒ x(x + 44) -33(x + 44) = 0
⇒ (x + 44)(x - 33) = 0
⇒ x = - 44, 33
Speed cannot be negative.
Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.
Exercise 4.3 Page Number 88

11

Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find 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 4yrespectively and their areas will be x2 and y2 respectively.
It is given that
4x - 4y = 24
x - y = 6
x = y + 6
Also, x2 + y2 = 468
⇒ (6 + y2) + y2 = 468
⇒ 36 + y2 + 12y + y2 = 468
⇒ 2y2 + 12y + 432 = 0
⇒ y2 + 6y - 216 = 0
⇒ y2 + 18y - 12y - 216 = 0
⇒ y(y +18) -12(y + 18) = 0
⇒ (y + 18)(y - 12) = 0
⇒ y = -18, 12
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.
Exercise 4.3 Page Number 88

1

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2x2 - 3x + 5 = 0
(ii) 3x2 - 4√3x + 4 = 0
(iii) 2x2 - 6x + 3 = 0

Answer

(i) Consider the equation
x2 - 3x + 5 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 2, b = -3 and c = 5
Discriminant = b2 - 4ac
= ( - 3)2 - 4 (2) (5) = 9 - 40
= - 31
As b2 - 4ac < 0,
Therefore, no real root is possible for the given equation.

(ii) 3x2 - 4√3x + 4 = 0
Comparing it with ax2 + bx + c = 0, we get
a = 3, b = -4√3 and c = 4
Discriminant = b2 - 4ac
= (-4√3)2 - 4(3)(4)
= 48 - 48 = 0
As b2 - 4ac = 0,
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be -b/2a and -b/2a.-b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and 2/√3.

 
(iii) 2x2 - 6x + 3 = 0

Comparing this equation with ax2 + bx + c = 0, we get
a = 2, b = -6, c = 3
Discriminant = b2 - 4ac

= (-6)2 - 4 (2) (3)
= 36 - 24 = 12
As b2 - 4ac > 0,
Therefore, distinct real roots exist for this equation:

Exercise 4.4 Page Number 91

2(i)

Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0

Answer

2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0, we get
a = 2, b = k and c = 3
Discriminant = b2 - 4ac

= (k)2 - 4(2) (3)
= k2 - 24
For equal roots,
Discriminant = 0
k2 - 24 = 0
k2 = 24
k = ±√24 = ±2√6
Exercise 4.4 Page Number 91

2(ii)

kx (x - 2) + 6 = 0

Answer

kx(x - 2) + 6 = 0
or kx2 - 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = k, b = - 2k and c = 6
Discriminant = b2 - 4ac
= ( - 2k)2 - 4 (k) (6)
= 4k2 - 24k
For equal roots,
b2 - 4ac = 0
4k2 - 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 'x2' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.
Exercise 4.4 Page Number 91

3

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Answer

Let the breadth of mango grove be l.
Length of mango grove will be 2l.
Area of mango grove = (2l) (l)= 2l2
2l2 = 800
l2 = 800/2 = 400
l2 - 400 =0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = 0, c = 400
Discriminant = b2 - 4ac
= (0)2 - 4 × (1) × ( - 400) = 1600
Here, b2 - 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 = 40 m
Exercise 4.4 Page Number 91

4

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.
then the age of the other friend will be (20 - x) years.
4 years ago,

Age of 1st friend = (x - 4) years
Age of 2nd friend = (20 - x - 4) = (16 - x) years
A/q we get that,
(x - 4) (16 - x) = 48
16x - x2 - 64 + 4x = 48
 - x2 + 20x - 112 = 0
x2 - 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -20 and c = 112
Discriminant = b2 - 4ac
= (-20)2 - 4 × 112
= 400 - 448 = -48
b2 - 4ac < 0
Therefore, there will be no real solution possible for the equations. Such type of condition doesn't exist.
Exercise 4.4 Page Number 91

5

Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

Answer

Let the length and breadth of the park be l and b.
Perimeter = 2 (l + b) = 80
l + b = 40
Or, b = 40 - l
Area = l×b = l(40 - l) = 40l - l240l -  l2 = 400
l2 -  40l + 400 = 0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = -40, c = 400
Discriminant = b2 - 4ac
(-40)2 - 4 × 400
= 1600 - 1600 = 0
b2 - 4ac = 0
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,l = -b/2a
l = (40)/2(1) = 40/2 = 20
Therefore, length of park, l = 20 m
And breadth of park, b = 40 - l = 40 - 20 = 20 m.
Exercise 4.4 Page Number 91

Ads