Class 10 Math Chapter 4 Exercise 4.2 NCERT Solution – Quadratic Equations

NCERT Class 10th Math chapter 4 Exercise 4.2 Question Answer. NCERT Solution for Class 10 Maths Chapter 4 Exercise 4.2 of Quadratic Equations NCERT Solution for CBSE, HBSE, RBSE and Up Board and all other chapters of class 10 math.

Also Read:- Class 10 Math NCERT Solution

NCERT Class 10 Math chapter 4 Quadratic Equations Exercise 4.2 Question Answer Solution.

Class 10 Math Chapter 4 Ex. 4.2 Question Answer


1. Find the roots of the following quadratic equations by factorization:
(i) x– 3x – 10 = 0

Ans. x– 3x – 10 = 0
x– 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
Therefore, the roots of x– 3x – 10 = 0 are 5 and -2.


(ii) 2x+ x – 6 = 0

Ans. 2x+ x – 6 = 0
2x+ 4x – 3x  – 6 = 0
2x(x + 2) -3(x + 2) = 0
(x + 2)(2x – 3) = 0
x + 2 = 0 or 2x – 3 = 0
x = -2, or x = 3/2
Therefore, the roots of 2x+ x – 6 = 0 are -2 and 3/2.


(iii) √2x+ 7x + 5√2 = 0

Ans. √2x+ 7x + 5√2 = 0
√2x+ 2x + 5x + 5√2 = 0
√2x(x + √2) + 5(x + √2) = 0
(x + √2)(√2x + 5) = 0
x + √2 = 0 or √2x + 5 = 0
x = – √2 or x = – 5/√2
Therefore, the roots of √2x+ 7x + 5√2 = 0 are – √2 and – 5/√2


(iv) 2x– x + 1/8 = 0

Ans. 2x– x + 1/8 = 0 ( multiplying whole equation to 8)
16x-8x + 1 = 0
16x-4x – 4x + 1 = 0
4x (4x – 1) – 1(4x – 1) = 0
(4x -1)(4x – 1) = 0
4x -1 = 0 or 4x -1 = 0
x = 1/4 or x = 1/4
Therefore, the roots of 2x– x + 1/8 = 0 are 1/4 and 1/4.


(v) 100x– 20x + 1 = 0

Ans. 100x– 20x + 1 = 0
100x– 10x – 10x + 1 = 0
10x(10x – 1) – 1(10x – 1) = 0
(10x – 1)(10x – 1) = 0
(10x – 1) = 0 or (10x – 1) = 0
x = 1/10 or x = 1/10
Therefore, the roots of 100x– 20x + 1 = 0 are 1/10 and 1/10.


2. Solve the problems given in example 1.

Example 1. Represent 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 Rs. 750. We would like to find out the number of toys produced on that day.

Ans. (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– 200 + 5x
As given that product = 124
– x2 + 45x – 200 = 124 (rearranging)
x– 45x + 324 = 0
[Now, we have to find 2 such number whose product is 324 and whose sum is -45. In this we will factorize 324 as 4 × 81 = 4 × 9 × 9. As you see we get 9 and 36. Now evaluate their signs correctly.

Note : Not to write in exam, just to understand students]
x– 9x – 36x + 324 = 0
x(x – 9) – 36(x – 9) = 0
(x – 36)(x – 9) =
On comparing equal to zero
we get x = 9, 36
If John had 9 marbles than Jivanti had 36 marbles and also if John had 36 marbles than Jivanti had 9 marbles.

(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in Rs ) of each toy that day = 55 – x
So, the total cost of production (in Rs) that day = x(55 – x)
Therefore, x(55 – x) = 750
55x – x2 = 750 (arranging)
x2 – 55x + 750 = 0
[Now, we have to find 2 such number whose product is 750 and whose sum is -55. In this we will factorize 750 as 10 × 75 = 10 × 3 × 25. As you see we get 30 and 25. Now evaluate their signs correctly.

Note : Not to write in exam, just to understand students]
x2 – 25x – 30x + 750 = 0
x(x – 25) – 30(x – 25) = 0
(x – 25)(x – 30) = 0
on comparing to zero,
x = 25, 30
So, if number of days is 25 then cost of production of each toy is 30 Rs and if number of days is 30 then the cost of production of each toy is Rs. 25.


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

Ans. Let x be the first number.
Then 2nd number will be 27 – x.
According to question,
x(27 – x) = 182
27x – x= 182 ( rearranging)
x– 27x + 182 = 0
[Now, we have to find 2 such number whose product is 182 and whose sum is -27. In this we will factorize 182 as 2 × 91 = 2 × 13 × 7. As you see we get 13 and 14. Now evaluate their signs correctly.

Note : Not to write in exam, just to understand students]
x2 – 13x – 14x + 182 = 0
x(x – 13) – 14(x – 13) = 0
(x – 13)(x- 14) = 0
On comparing
x – 13 = 0
x = 13
Or x – 14 = 0
x = 14
From there we get that the required two number are x and 27 – x. If first number is 13 then 2nd will be 14 and if 1st number is 14 then second number will be 13.
Therefore, required numbers are 13 and 14.


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

Ans. Let x be the first positive integer than second consecutive integer will be x + 1.
According to question,
x+ (x + 1)= 365
x2 + x2 + 1 + 2x = 365
2x2 + 2x + 1 = 365 ( rearranging)
2x2 + 2x + 1 – 365 = 0
2x2 + 2x – 364 = 0 ( divide equation by 2)
x2 + x – 182 = 0
[ Now, we have to find 2 such numbers whose product is – 182 and whose sum is 1. In this we will factorize 182 as 2 × 91 = 2 × 13 × 7 . As you see we get 13 and 14. Now evaluate their signs correctly

Note : Not to write in exam, just to understand students]
x2 +14x – 13x – 182 = 0
x(x + 14) -13(x – 14) = 0
(x – 13)(x + 14) = 0
On comparing
x – 13 = 0
x = 13
and x + 14 = 0
x = -14, as it is not positive therefore it can’t be such number.
Therefore, x = 13 and x +1 = 14
The required numbers are 13 and 14.


5. The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.

Ans. Let x be the base of right angle triangle. Then altitude will be x – 7.
According to question,
Using pythagoras theorem,
x2 + (x – 7)= 132
x+ x+ 49 – 14x = 169 (rearranging)
2x– 14x + 49 – 169 = 0
2x2 – 14x – 120 = 0 ( dividing equation by 2)
x– 7x – 60 = 0
[ Now , we have to find 2 such numbers such that their product is -60 and their sum is -7. In this we will factorizing 60 = 2 × 2 ×15 = 2 × 2 × 3 × 5. As you see we get 12 and 5. Now evaluate their signs correctly.

Note : Not to write in exam, just to understand students]
x2 – 12x + 5x – 60 = 0
x(x – 12) + 5(x – 12) = 0
(x – 12)(x + 5) = 0
On comparing equal to zero, we get
x – 12 = 0 or x + 5 = 0
x = 12 or x = -5
As side of any triangle cannot be negative. Therefore, value of x will be 12.
If x = 12 than x – 7 will be 5.
Therefore, base of given right angle triangle is 12 cm and altitude of the triangle is 5 cm.


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.

Ans. Let x be the number of article produced each day. Then cost of production of production of each article will be 2x + 3 Rs.
According to question,
x(2x + 3) = 90
2x2 + 3x – 90 = 0
[ Now , we have to find 2 such numbers such that their product is -180 and their sum is + 3. In this we will factorizing 180 = 9 × 20 = 3 × 3 × 4 × 5. As you see we get 12 and 15. Now evaluate their signs correctly.

Note : Not to write in exam, just to understand students]
2x– 12x + 15x – 90 = 0
2x(x – 6) + 15(x – 6) = 0
(x – 6)(2x + 15) = 0
On comparing equal to zero, we get
x – 6 = 0 or 2x + 15 = 0
x = 6 or -15/2
As we cannot take negative value of x.
Therefore, value of x which is number of article is 6 and cost of production of each article is 2x + 3 = 15 Rs.


 

Leave a Comment

error: