Master Class 10 Math Chapter 4 Quadratic Equations Ex 4.2

In Exercise 4.2 of Class 10 Mathematics, we focus on solving quadratic equations using the factorization method. This method involves expressing a quadratic equation as a product of two linear factors and finding the roots by setting each factor equal to zero.

The general form of the quadratic equation we work with is:
ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0

By factorizing these equations, we can find the values of x, also known as the roots of the equation. The factorization method is based on splitting the middle term into two terms, which allows us to group and simplify the equation.

In this exercise, we will apply the factorization method to different types of quadratic equations and find their solutions. Let’s get started!

Class 10 Math Chapter 4 Quadratic Equations 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 + [Math Processing Error] = 0 8
(v) 100 x² – 20 X + 1 = 0
Solution:

(i) x² − 3x − 10 = 0
We need two numbers whose product = −10 (constant term) and sum = −3 (coefficient of x).
Numbers = −5 and +2.

So,
x² − 3x − 10 = x² − 5x + 2x − 10
= x(x − 5) + 2(x − 5)
= (x − 5)(x + 2)

⇒ Roots: x = 5, −2

(ii) 2x² + x − 6 = 0
Product = 2 × (−6) = −12, Sum = +1
Numbers = +4 and −3.

So,
2x² + x − 6 = 2x² + 4x − 3x − 6
= 2x(x + 2) − 3(x + 2)
= (2x − 3)(x + 2)

⇒ Roots: x = 3/2, −2

(iii) √2x² + 7x + 5√2 = 0
Product = √2 × 5√2 = 10, Sum = 7
Numbers = 5 and 2.

So,
√2x² + 7x + 5√2 = 0

= √2x² + 5x + 2x + 5√2 =0
= x(√2x + 5) + √2(√2x + 5) = 0
= (√2x + 5)(x + √2) = 0

⇒ Roots: x = −5/√2, −√2

(iv) 2x² − x + 1/8 = 0
Multiply by 8 to remove fraction:
16x² − 8x + 1 = 0

Now, this is a perfect square trinomial:
16x² − 8x + 1 = (4x − 1)²

So,
(4x − 1)² = 0
⇒ x = 1/4 (double root)

(v) 100x² − 20x + 1 = 0
This is also a perfect square:
100x² − 20x + 1

= (10x − 1)²
= (10x − 1)² = 0
⇒ x = 1/10 (double root)

Question 2.
Solve 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 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 ₹750. We would like to find out the number of toys produced on that day.
Solution:
(i) John and Jivanti
Let John had x marbles and Jivanti had y marbles.
Given:
x + y = 45 … (1)
After each loses 5: (x − 5)(y − 5) = 124 … (2)

From (1): y = 45 − x. Substitute in (2):
(x − 5)[(45 − x) − 5] = 124
(x − 5)(40 − x) = 124
−x² + 45x − 200 = 124
−x² + 45x − 324 = 0
x² − 45x + 324 = 0
Factor: (x − 9)(x − 36) = 0
So x = 9 or 36.
Then y = 36 or 9.

Answer: They had 9 and 36 marbles originally.

(ii) Toys produced in a day
Let the number of toys be n.
Cost per toy = 55 − n.
Total cost = n(55 − n) = 750.

So:
−n² + 55n − 750 = 0
n² − 55n + 750 = 0
Factor: (n − 25)(n − 30) = 0
So n = 25 or n = 30.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the numbers be x and y.

Given:
x + y = 27 and xy = 182.

Form the quadratic with sum and product:
t² − (sum)t + (product) = 0 ⇒ t² − 27t + 182 = 0.

Factorise:
t² − 27t + 182 = (t − 13)(t − 14) = 0
⇒ t = 13 or 14.

So, the two numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, the sum of whose squares is 365.
Solution:
Let the two consecutive positive integers be x and x + 1.

According to the question,
x² + (x + 1)² = 365

⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x + 1 = 365
⇒ 2x² + 2x − 364 = 0
⇒ x² + x − 182 = 0

Now, factorise:
x² + x − 182 = (x + 14)(x − 13) = 0

So, x = −14 or x = 13.

Since we need positive integers,
x = 13.

Therefore, the 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:
Solution:
Let the base of the right triangle be x cm.
Then altitude = (x − 7) cm.
Hypotenuse = 13 cm.

By Pythagoras theorem:
(base)² + (altitude)² = (hypotenuse)²

x² + (x − 7)² = 13²
x² + (x² − 14x + 49) = 169
2x² − 14x + 49 = 169
2x² − 14x − 120 = 0
Divide by 2: x² − 7x − 60 = 0

Factorise:
x² − 7x − 60 = (x − 12)(x + 5) = 0
So, x = 12 or −5.

Since base cannot be negative,
x = 12.

Then altitude = x − 7 = 12 − 7 = 5.

The base = 12 cm and the altitude = 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 ₹90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced be x.

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

Total cost of production = (number of articles) × (cost per article)
= x(2x + 3) = 90

⇒ 2x² + 3x − 90 = 0
Divide by common factor: (no common factor)

Factorise:
2x² + 15x − 12x − 90 = 0
⇒ (2x² + 15x) − (12x + 90) = 0
⇒ x(2x + 15) − 6(2x + 15) = 0
⇒ (x − 6)(2x + 15) = 0

So, x = 6 or x = −15/2.

Since number of articles must be positive, x = 6.

Now, cost per article = 2x + 3 = 2(6) + 3 = 15.

The number of articles produced = 6,
Cost of each article = ₹15.