Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.2

Question 1.

Find the roots of the following quadratic equations by factorisation :

(i) x^{2} – 3x – 10 = 0

(ii) 2x^{2} + x – 6 = 0

(iii) \(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0

(iv) 2x^{2} – x + \(\frac{1}{8}\) = 0

(v) 100x^{2} – 20x + 1 = 0

Solution:

(i) According to the question the quadratic equation is

x^{2} – 3x – 10 = 0 | S = – 3

or x^{2} – 5x + 2x – 10 = 0 | P = – 10

or x (x – 5) + 2 (x – 5) = 0

or (x – 5) (x + 2) = 0

i.e., x – 5 = 0 or x + 2 = 0

x = 5 or x = – 2

Hence, 5 and – 2 are the roots of the given quadratic equation.

(ii) According to the question the quadratic equation is

2x^{2} + x – 6 = 0 S = 1

or 2x^{2} + 4x – 3x – 6 = 0 P = -6 × 2

or 2x(x + 2) – 3(x + 2) = 0 = -12

(x + 2) (2x — 3) = O

i.e., x + 2 = 0 or 2x – 3 = 0

x = -2 or x = \(\frac{3}{2}\)

Hence, -2 and \(\frac{3}{2}\) are the roots of the given quadratic equation.

(iii) According to the question quadratic equation is—

\(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0 | S = 7

or \(\sqrt{2}\)x^{2} + 2x + 5x + 5\(\sqrt{2}\) = 0 | P = \(\sqrt{2}\) × 5\(\sqrt{2}\)

or \(\sqrt{2}\)x(x + \(\sqrt{2}\)) + 5(x + \(\sqrt{2}\)) = 0 = 10

or (x + \(\sqrt{2}\))(\(\sqrt{2}\)x + 5) = 0

i.e., x + \(\sqrt{2}\) = 0 or \(\sqrt{2}\)x + 5 = 0

x = –\(\sqrt{2}\) or x = \(\frac{-5}{\sqrt{2}}\)

Hence, –\(\sqrt{2}\) and \(\frac{-5}{\sqrt{2}}\) are the roots of the given quadratic equation.

(iv) According to the question the quadratic equation is

2x^{2} – x + \(\frac{1}{8}\) = 0

or \(\frac{16 x^{2}-8 x+1}{8}\) = 0 | S = -8

or 16x^{2} – 8x + 1 = 0 | P = 16 × 1 = 16

or 16x^{2} – 4x – 4 + 1 = 0

or 4x(4x – 1) – 1(4x – 1) = 0

i.e., 4x – 1 = 0

or 4x – 1 = 0

x = \(\frac{1}{4}\) or x = \(\frac{1}{4}\)

Hence, \(\frac{1}{8}\) and \(\frac{1}{8}\) are the roots of the given quadratic equation.

(v) According to the question the quadratic equation is are the roots of the given quadratic equation.

100x^{2} – 20x + 1 = 0

or 100x^{2} – 10x – 10x + 1 = 0 | S = – 20

(10x – 1) 10x -1) = 0 | P = 100 × 1 = 100

(10x – 1) (10x – 1) = 0

10x – 1 = 0 or 10x – 1 = 0

x = \(\frac{1}{10}\) or x = \(\frac{1}{10}\)

Hence, \(\frac{1}{10}\) and \(\frac{1}{10}\) are the roots of the given quadratic equation.

Question 2.

Solve the problems given in Example

1. The statements of these problems are given below :

(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 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 produc¬tion was ? 750. We would like to find out the number of toys produced on that day.

Solution:

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

Then the number of marbles Jivanti had = 45 – x

The number of marbles left with John, when he lost 5 marbles = x – 5

The number of marbles left with Jivanti, when she lost 5 marbles

= 45 – x – 5

= 40 – x

Therefore,

their product = (x – 5)(40 – x)

= 40x – x^{2} – 200 + 5x

= -x^{2} + 45x – 200

According to the question—

– x^{2} + 45x – 200 = 124

or – x^{2} + 45x – 324 = 0 | S = – 45,

or x^{2} – 45x + 324 = 0 | P = 324

or x^{2} – 36x – 9x + 324 = 0

or x (x – 36) – 9 (x – 36) = 0

or (x – 36) (x – 9) = 0

i.e., x – 36 = 0 or x – 9 = 0

x = 36 or x = 9

∴ x = 36, 9

Therefore, the number of marbles they had to start with is 36 and 9 or 9 and 36.

(ii) Let the number of toys produced on that day = x

Therefore, the cost of production (in rupees) of each toy that day = 55 – x

So, the total cost of production (in rupees) that day

= x (55 – x)

According to the question

x (55 – x) = 750

or 55x – x^{2} = 750

or -x^{2} + 55x – 750 = 0 | S = – 33,

or x^{2} – 55x + 750 = 0 | P = 750

or x^{2} – 30x – 25x + 750 = 0

or x(x – 30) – 25 (x – 30) = 0

or (x – 30) (x – 25) = 0

i.e., x – 30 = 0 or x – 25 = 0

x = 30 or x = 25

∴ x = 30, 25

Hence, the number of toys produced on that day is 30 and 25 or 25 and 30.

Question 3.

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

Solution:

Let first number = x

Then second number = 27 – x

Hence, their product = x (27 – x)

= 27x – x^{2}

According to the question—

27x – x^{2} = 182

or -x^{2} + 27x – 182 = 0 |S = -27

or x^{2} – 27x + 180 = 0 |P = 182

or x^{2} – 13x – 14x + 180 = 0

or x(x – 13) – 14(x – 13) = 0

or (x – 13)(x – 14) = 0

i.e., x – 13 = 0 or x – 14 = 0

x = 13 or x = 14

∴ x = 13, 14

Hence, the two numbers are 13 and 14 or 14 and 13.

Question 4.

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

Solution:

Let the first positive integer = x

Then,the other positive integer = x + 1

According to the question

(x)^{2} + (x + 1)^{2} = 365

or x^{2} + x^{2} + 1 + 2x = 365

or 2x^{2} + 2x + 1 – 365 = 0

or 2x^{2} + 2x – 364 = 0

or x^{2} + x – 182 = 0

or x^{2} + 14x – 13x – 182 = 0

or x (x + 14) – 13(x + 14) = 0

or (x + 14) (x – 13) = 0

i.e., x + 14 = 0 or x – 13 = 0

x = – 14 or x = 13

∵ We want positive integer

So x = – 14 is not possible

∴ x = 13

∴ First positive integer = 13

and second positive integer = 13 + 1 = 14

Hence, the required two consecutive positive integers are 13 and 14.

Question 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.

Solution:

Let the base of the right triangle = x cm.

Then, the altitude (perpendicular) of the right triangle

= (x – 7) cm.

The hypotenuse of the right triangle = 13 cm. (Given)

According to Pythogoras Theorem,

(Base)^{2} + (Perpendicular)^{2} = (Hypotenuse)^{2}

or (x)^{2} + (x – 7)^{2} = (13)^{2}

or x^{2} + x^{2} + 49 – 14 x = 169

or 2x^{2} – 14x + 49 – 169 = 0

or 2x^{2} – 14x – 120 = 0

or 2 [x^{2} – 7x – 60] = 0

or x^{2} – 7x – 60 = 0 | S = – 7

or x^{2} – 12x + 5x – 60 = 0 | P = – 60

or x (x – 12) + 5 (x – 12) = 0

or (x – 12) (x + 5) = 0

i.e., x – 12 = 0 or x + 5 = 0

x = 12 or x = – 5

∵ The length of the base of the triangle can never be negative

So, discarding x = – 5

∴ x = 12

So, Base of the right triangle = 12 cm and, Altitude (perpendicular) of the right triangle

= (12 – 7) cm.

= 5 cm.

Question 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.

Solution:

Let the number of articles produced in a day = x

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

∴ The total cost of production on a particular day

= Rs. [x (2x + 3)]

= Rs. (2x^{2} + 3x)

According’to the question,

2x^{2} + 3x = 90

or 2x^{2} + 3x – 90 = 0 | S = 3,

P = 2 × -90 = -180

or 2x^{2} – 12x + 15x – 90 = 0

or 2x (x – 6) + 15 (x – 6) = 0

or (x – 6) (2x + 15) = 0

i.e., x – 6 = 0 or 2x + 15 = 0

x = 6 or x = \(\frac{-15}{2}\)

∵ The number of articles cannot be negative

Therefore leaving x = \(\frac{-15}{2}\)

∴ x = 6

Hence, the number of articles produced on a particular day = 6

and the cost of production of each article = Rs. [2 × 6 + 3]

= Rs. 15

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