# KSEEB Solutions for Class 10 Maths Chapter 10 Quadratic Equations Ex 10.2

Students can Download Class 10 Maths Chapter 10 Quadratic Equations Ex 10.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 10 Quadratic Equations Ex 10.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) $$\sqrt{2}$$x2 + 7x + 5$$\sqrt{2}$$ = 0
(iv) 2x2 – x + $$\frac{1}{8}$$ = 0
(v) 100 x2 – 20x + 1 = 0
i) x2 – 3x – 10 = 0
x2 – 5x + 2x – 10 = 0
x (x – 5) + 2 (x – 5) = 0
(x – 5) (x + 2) = 0
x – 5 = 0 (or) x + 2 = 0
x = 5 (or) x = – 2
∴ x = 5, x = – 2
∴ 5 & – 2 are the roots of the equation x2 – 3x – 10 = 0

ii) 2x2 + x – 6 = 0

2x2 + 4x – 3x – 6 = 0
2x (x + 2) – 3 (x + 2) = 0
(x + 2) (2x – 3) = 0
x + 2 = 0 (or) 2x – 3 = 0
x = $$\frac{3}{2}$$
x = – 2 (or) 2x = 3
x = $$\frac{3}{2}$$
x = – 2, $$\frac{3}{2}$$
∴ – 2 & $$\frac{3}{2}$$ are the roots of the equation 2x2 + x – 6 = 0

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

(iv) 2x2 – x + $$\frac{1}{8}$$ = 0
16x2 – 8x + 1 = 0
16x2 – 4x – 4x + 1 = 0
4x(4x – 1) – 1(4x- 1) = 0
(4x – 1) (4x – 1) = 0
(4x – 1)2 = 0
If 4x – 1 = 0, then 4x = 1
∴ x = $$\frac{1}{4}$$
∴ Two roots are $$\frac{1}{4}, \frac{1}{4}$$

v) 100x2 – 20x + 1 =0

100x2 – 10x – 10x + 1 =0
10x(10x – 1) – 1(10x – 1) = 0
(10x – 1)(10x – 1) = 0
10x – 1 = 0 (or) 10x – 1 = 0
10x = 1 or 10x = 1
x = $$\frac{1}{10}$$ (or) x = $$\frac{1}{10}$$
x = $$\frac{1}{10}$$, $$\frac{1}{10}$$
∴ $$\frac{1}{10}$$ & $$\frac{1}{10}$$ are the roots of the equation of 100x2 – 20x + 1 =0

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.
Let the number of marbles John had be ‘x’ & Jivanti had be 45 – x
John lost 5 marbles = (x – 5) & Jivanti lost 5 marbles = 45 – x – 5 = 40 – x.
their product = (x – 5) (40 – x) = 124
40x – 200 – x2 + 5x = 124
– x2 + 45x – 200 – 124 = 0
– x2 + 45x – 324 = 0
Multiply by ‘_’

x2 – 45x + 324 = 0
x2 – 36x – 9x + 324 = 0
x (x – 36) – 9 (x – 36) = 0
(x – 36) (x – 9) = 0
x – 36 = 0 or x – 9 = 0
x = 36 (or) x = 9
x = 36, 9
∴ Therefore John and Jivanti had 36 and 9 (or) 9 and 36 marbles.

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.
Let the number of toys produced on that day be ‘x’
The cost of production of each toy = 55 – x
The total cost of Production that day
= x (55 – x)
∴ x (55 – x) = 750
55x – x2 = 750
x2 – 55x + 750 = 0

x2 – 30x – 25x + 750 = 0
x (x – 30) – 25 (x – 30) = 0
(x – 30) (x – 25) = 0
x – 30 = 0 or x – 25 = 0
x = 30 (or) x = 25
The number of toys produced on that day has 25 (or) 30.

Question 3.
Find two numbers whose sum is 27 and the product is 182.
In those numbers, let one of the numbers be ‘x’.
Another number is (27 – x).
Their product is 182.
∴ x (27 – x) = 182
27x – x2 = 182
-x2 + 27x – 182 = 0
x2 – 27x + 182 = 0
x2 – 14x – 13x + 182 = 0
x(x – 14) – 13(x – 14) = 0
(x – 14) (x – 13) = 0
If x – 14 = 0, then x = 14
If x – 13 = 0, then x = 13
∴ x = 14 OR x = 13
∴ The Numbers are 14 and 13.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Let two consecutive Positive integers be x and x + 1
Sum of their squares = 365
x2 + (x + 1)2 = 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 (x + 14) (x – 13) = 0
x + 14 = 0 (or) x – 13 = 0
x = – 14 or x = 13
x= 13
Two consecutive numbers are x, x + 1
∴ 13, 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.
Let the base of right angle triangle be ‘x’ and altitude be x – 7, hypotenuse = 13 cm
(hypotenuse)2 = (base)2 + (Alitutude)2 [By Pythagoras theorem]
(13)2 = x2 + (x – 7)2
169 = x2 + x2 – 14x + 49
2x2 – 14x +49 – 169 = 0
2x2 – 14x – 120 = 0 Divide by 2
x2 – 7x – 60 = 0
x2 – 12x + 5x – 60 = 0
x(x – 12) + 5 (x – 12) = 0
(x – 12)(x + 5) = 0
x – 12 = 0 (or) x + 5 = 0
x = 12 or x = – 5
x = 12
∴ base of Right angle triangle
= x = 12 cm
∴ Altitude of Right angle triangle
= x – 7 = 12 – 7 = 5 cm
∴ Other two sides are 12 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.
Let the number of pottery articles be ‘x’.
Cost of production of each (in rupees) is = (2x + 3)
∴ 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
If x – 6 = 0, then x = 6
If 2x + 15 = 0, then 2x = -15
∴ x = $$\frac{-15}{2}$$
∴ Number of pottery articles produced is 6.
Cost of production of 6 pottery articles is Rs. 90
Cost of production of 1 pottery article ……….. ?
$$\frac{1 \times 90}{6}$$ = Rs. 15
OR Cost of Production= 2x + 3
= 2 × 6 + 3
= 12 + 3
= Rs. 15.