NCERT Solutions For Class 10 Math Chapter – 4 Exercise – 4.2
Q1. Find the roots of the following quadratic equations by factorisation:
- X2– 3x – 10 = 0
- 2x2+ x – 6 = 0
- √2x2+ 7x + 5√2 = 0
- 2x2– x + 1/8 = 0
- 100x2– 20x + 1 = 0
Solution:-
- X2– 3x – 10
= x2 – 5x + 2x – 10
= x(x-5) + 2(x-5)
= (x-5) (x +2)
X = -5 or x = 2
- 2x2+ x – 6
= 2x2 + 4x – 3x – 6
= 2x(x+2) -3(x+2)
= (x+2)(x-3)
X = -2 or x = 3/2
- √2x2+ 7x + 5√2
= √2x2 + 5x + 2x + 5√2
= x(√2x +5) + √2(√2x + 5)
= x = -5/√2 or x = -√2 .
- 2x2– x + 1/8
= 1/8 (16x2– 8x + 1)
= 1/8 ( 16x2 – 4x – 4x +1)
= 1/8 (4x – 1)2
(4x – 1)(4x – 1)
X = 1/4 or x = 1/4 .
- 100x2– 20x + 1
= 100x2 – 10x – 10x + 1
= 10x( 10x – 1) – 1(10x – 1)
= (10x – 1)2
(10x – 1)(10x – 1)
X = 1/10 or x = 1/10
Q2. (I) John and Jivanti together have 45 marbles Both of them lots 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.
Solution:-
(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
X = 36 or x =9
(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.
Solution:-
X(55-x) = 750
= x2 – 55x + 750 = 0
= x2 – 25x – 30x + 750 = 0
X(x-25) -30(x-25) = 0
X = 25 or x = 30
Q3. Find two number whose sum is 27 and product is 182.
Solution:-
X(27 – x) = 182
= x2 – 13x – 14x + 182 = 0
= x(x-13) – 14(x – 13) = 0
(x-13)(x-14) = 0
X = 13 or x = 14
First number = 13
Other number = 27 – 13 = 14
First number = 14
Other number = 27 – 14 = 13
The number are 13 and 14.
Q4. Find two consecutive positive integer , sum of whose square is 365.
Solution:-
X2 + (x + 1)2 = 365
= x2 + x2 + 1 + 2x = 365
= 2x2 + 2x – 364 = 0
= x2 + x – 182 = 0
= x2 + 14x – 13x -182 = 0
= x(x+14) – 13(x + 14) = 0
X = 14 and x = 13
Q5. 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:-
X2 + (x – 7)2 = 132
= x2 + x2 + 49 – 14x = 169
= 2x2 – 14x – 120 = 0
= X2 – 7x – 60 = 0
= (x-12)(x+5) = 0
= x = 12 or x = -5
Q6. 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:-
2x2 + 3x – 90 = 0
2x2+ 15x – 12x – 90 = 0
X(2x + 15) -6(2x +15) = 0
(2x +1 5)(x – 6) = 0
X = -15/2 or x = 6