NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

These NCERT Solutions for Class 10 Maths Chapter Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2

NCERT Solutions for Class 10 Maths Chapter Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2x2 + 7x + 5√2 = 0
(iv) 2x2 – x + \(\frac { 1 }{ 8 }\) = 0 8
(v) 100x2 – 20 x + 1 = 0
Solution:
(i) We have,
x² – 3x – 10 = 0
or x² – 5x + 2x – 10 = 0
or x(x – 5) – 2(x – 5) = 0
or, (x – 5) (x + 2) = 0
∴ x – 5 = 0 or, x + 2 = 0
⇒ x = 5 or, x + 2 = 0
Therefore, the roots x² – 2x -10 = 0 are – 5 and

(ii) We have,
2x² + x – 6 = 0
or, 2x² + 4x – 3x – 6 = 0
or, 2x(x + 2) – 3(x + 2) = 0
or, (x + 2) (2x – 3) = 0
∴ (x + 2) = 0 or, (2x – 3) = 0
⇒ x = – 2 or, x = \(\frac { 3 }{ 2 }\)
Therefore, the roots of 2x² + x – 6 = 0 are – 2 and \(\frac { 3 }{ 2 }\)

(iii) We have,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 1
Therefore, the root of \(\sqrt{2}\)2x² + 7x + 5\(\sqrt{2}\) = o are \(\frac{-5}{\sqrt{2}}\) and \(\sqrt{2}\)

(iv) We have,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 2
Therefore, the roots of 2x² – x + \(\frac { 1 }{ 8 }\) = 0 is \(\frac { 1 }{ 4 }\) ; \(\frac { 1 }{ 4 }\)

(v) We have,
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 3
Therefore, the roots of 100x² – 20x + 1 = 0 is \(\frac { 1 }{ 10 }\), \(\frac { 1 }{ 10 }\).

NCERT Solutions for Class 10 Maths Chapter Ex 4.2

Question 2.
Solve the problems given in Example 1.
(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 (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 ₹ 750. We would like to find out the number of toys produced on that day.
Solution:
(i) Let number of marbles John have be x
∴ Number of marbles Jivanti have = 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 = 45x – x – 5 = 40 – x
According to question,
(x – 5) (40 – x) = 124
or 40x – x² – 200 + 5x = 124
or 45x – x² – 200 = 124
or x² – 45x + 324 = 0
or x² – 36x – 9x + 324 = 0
or x(x – 36) – 9(x – 36) = 0
or (x – 36) (x – 9) = 0
∴ (x – 36) =0 or (x – 9) = 0
⇒ x = 36 or x = 9
Therefore, if number of marbles John have be 36
Then number of marbles Jivanti have = 45 – x = 45 – 36 = 9
And, if number of marbles John have be 9
Then, number of marbles Jivanti have = 45 – x = 45 – 9 = 36

(ii) Let the number of toys produced on that day be 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 question,
x(55 – x) = 750
or 55x – x² = 750
or x² – 55x + 750 = 0
or x² – 30x – 25x + 750 = 0
or x (x – 30) – 25 (x – 30) = 0
(x – 30) (x – 25) = 0
∴ (x – 30) = 0 or (x – 25) = 0
⇒ x = 30 or x = 25
Therefore, if the number of toys produced on that day is x = 30, then, the cost of production (in rupees) or each toy = 55 – x = 55 – 30 = ₹ 25.
And, if the number of toys produced is x = 25,
then, the cost of production = 55 – x = 55 – 25 = ₹ 30.

Question 3.
Find two numbers whose sum is 27, and product is 182.
Solution:
Let first number be x Second number be = 27 – x According to question,
x(27 – x) = 182 ‘
or 27x – x² = 182
or x² – 27x + 182 = 0
or x² – 14x – 13 + 182 = 0
or x (x – 14) – 13 (x – 14) = 0
or (x – 14) (x – 13) = 0
∴ (x – 14) = 0 or, (x – 13) = 0
⇒ x = 14 or x = 13
Therefore, if first number is 14 then second number is
27 – x = 27 – 14 = 13
And, if first number is 13 then second number is
27 – x = 27 – 13 = 14

NCERT Solutions for Class 10 Maths Chapter Ex 4.2

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let first consecutive positive integer be x
∴Second consecutive positive integer be x + 1
According to question,
x² + (x + 1)² = 365
or x² + x² + 2x + 1 = 365
or 2x² + 2x + 1 – 365 = 0
or 2x² + 2x – 364 = 0
or 2(x² + x – 182) = 0
or x² + x – 182 = 0
or x² + 14x – 13x – 182 = 0
or x(x + 14) – 13 (x + 14) = 0
or (x + 14) (x- 13) = 0
∴ (x + 14) = 0 or (x – 13) = 0
⇒ x = – 14 or x = 13
But, we has given that numbers are positive
∴ x = – 14 is neglected.
Therefore, x = 13
∴ First consecutive positive integer = x = 13
and second consecutive positive integer = 27 – x = 14

Question 5.
The altitude of a right traingle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of right angle triangle = x
∴ The height of right angle triangle = x – 7
And hypotenuse of right angle triangle = 13 cm
By Pythagoras theorem, we have,
(height)² + (base)² = (hypotenuse)²
∴ (x – 7)² + x² = (13)²
or x² – 14x + 49 + x² = 169
or 2x² – 14x + 49 – 169 = 0
or 2x² -14x – 120 = 0
or 2(x² – 7x – 60) = 0
or x² – 7x – 60 = 0
or x² – 12x + 5x – 60 = 0
or x (x – 12) + 5 (x – 12) = 0
or (x -12) (x + 5) = 0
∴ x – 12 = 0 or x + 5=0
⇒ x = 12 or x = – 5
But, length cannot be in negative
So, x = – 5 is neglected.
∴ x = 12
Therefore, base of the right angle triangle = x
∴ x = 12 cm
and height of the right angle triangle = x – 7 = 12 – 7 = 5 cm

NCERT Solutions for Class 10 Maths Chapter Ex 4.2

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 numbr of articles produced on that day. It the total cost of production on that was ₹ 90. Find the number of articles produced and the cost of each article.
Solution:
Let total number of articles produced = x
Cost of production = 2x + 3
According to question,
x (2x + 3) = 90
or 2x² + 3x = 90
or 2x² + 3x – 90 = 0
or 2x² + 15x – 12x – 90 = 0
or x(2x + 15) – 6 (2x + 15) = 0
or (2x +15) (x – 6) = 0
∴(2x + 15) = 0 or (x – 6) = 0
x = \(\frac { -15 }{ 2 }\) or x = 6
But, number of articles cannot be in negative.
So, x = \(\frac { -15 }{ 2 }\) is neglected.
∴ Total number of articles produced = x = 6
and cost of production = 2x + 3 = 2 x 6 + 3 = ₹ 15.

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