NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations
Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations. This Exercise contains 6 questions, for which detailed answers have been provided in this note. Quadratic Equations exercise 4.2 class 10 chapter 4 NCERT Solutions have been explained in a simple and easy-to-understand language to help you learn and prepare for your upcoming class 10 Maths exams. Here we are sharing Quadratic Equations exercise 4.2 class 10 solution.
Category | NCERT Solutions for Class 10 |
Subject | Maths |
Chapter | Chapter 4 |
Exercise | Exercise 4.2 |
Chapter Name | Quadratic Equations |
NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations
You can also download the free PDF of NCERT Solutions Chapter 4 Ex 4.2 Quadratic Equations for your online exam preparation.
1. Find the roots of the following quadratic equations by factorization:
(i) x2-3x-10 = 0
Solution:
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
Hence the required roots are 5, -2.
(ii) 2x2+x-6=0
Solution:
2x2+x-6 = 0
2x2+4x-3x-6 = 0
2x(x+2)-3(x+2) = 0
(2x-3)(x+2) = 0
⇒ 2x-3 = 0 or, x+2 = 0
x = 3/2 or x = -2
Hence the required roots are 3/2, -2.
(iii) √2x2+7x+5√2 = 0
Solution:
√2x2+7x+5√2 = 0
√2x2+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)
Hence the required roots are -√2, -(5/√2).
(iv) 2x2-x+(1/8)Â =0
Solution:
2x2-x+ (1/8) =0
16x2-8x+1 = 0
(4x)2-2.4x.1+12 = 0
(4x-1)2= 0
⇒ 4x-1 = 0  or 4x-1 = 0
x = (1/4)Â or x = (1/4)
Hence the required roots are 1/4, 1/4
(v) 100x2-20x+1 = 0
Solution:
100x2-20x+1 = 0
(10x)2-2×10x.1+12= 0
(10x-1)2 = 0
10x-1 = 0 or 10x-1 = 0
x = 1/10 or x = 1/10
Hence the required roots are 1/10 or 1/10
2. Solve the problem given in example 1.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each others, and product of the marbles they now have is 124. We would like to find how many marbles they had to start with?
Solution: Let the number of marbles John had be x.
Then the number of marbles Jivanti had be 45-x
They lost 5 marbles to each others.
Hence now the number of marbles of John is x-5 and that of Jivanti is 45-x-5 = 40-x.
Then by given condition we have
(x-5)(40-x) = 124
40x-200-x2+5x = 124
x2-45x+324 = 0
x2-36x-9x+324 = 0
x(x-36)-9(x-36) = 0
(x-36)(x-9) = 0
x-36 = 0
x = 36
or x-9 = 0
x = 9
Therefore, they had 36 and 9 marvels respectively.
(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: Let the number of toys produced on that day is x.
Therefore, the cost of production of each toy on that day is (55-x)
So, the total cost of production on that day is = x(55-x)
Then using the given condition, we have,
x(55-x) = 750
55x-x2 =750
x2-55x+750 = 0
x2-25x-30x+750 = 0
x(x-25)-30(x-25) = 0
(x-25)(x-30) = 0
x-25 = 0
x =25
or x-30 =0
x = 30
Therefore, the number of toys produced on that day is 25 or 30.
3. Find the two numbers whose sum is 27 and the product is 182.
Solution: Let one number is x. Then another number is 27-x.
Then by the given condition,
x(27-x) = 182
27x- x2-182 = 0
x2-27x+182 = 0
x2-14x-13x+182 = 0
x(x-14)-13(x-14) = 0
(x-14)(x-13) = 0
x-14 = 0
x =14
or, x-13 = 0
x =13
Hence the required two numbers are 13 and 14.
4. Find the two consecutive positive integers, the sum of whose squares is 365.
Solution: Let the two consecutive positive integers be x and x+1
Then by the given condition,
x2+(x+1)2 = 365
x2+ x2+2x+1 = 365
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 or 13
Since the numbers are positive so x = -14 not possible.
So, the numbers are 13 and 13+1 = 14.
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: Let base of the right triangle is x cm. Then the altitude is x-7 cm.
Then by the property of right triangle and using the given property we have
x2+(x-7)2 = (13)2
x2+ x2-14x+49 = 169
2x2-14x-120 = 0
x2-7x-60 = 0
x2-12x+5x-60 = 0
x(x-12)+5(x-12) = 0
(x-12)(x+5) = 0
x-12 = 0
x =12
or, x+5 = 0
x = -5
Since length can’t be negative, so x = -5 is not possible.
Hence the base is 12 cm and the altitude is (12-7)cm = 5 cm
6. A cottage industry produces a certain number of pottery articles in a day. It was observed that 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 number 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 on that day is x. then the price on that day is 2x+3
Then using the given condition we have,
x(2x+3) = 90
2x2+3x-90 = 0
2x2+15x-12x-90 = 0
x(2x+15)-6(2x+15)=0
(2x+15)(x-6) = 0
2x+15 = 0
x = – (15/2)
or, x-6 = 0
x = 6
Since Articles can’t be negative then x = -(15/2) is not possible.
Therefore, the number of articles produced on that day = 6
Cost of each article = 2×6+3 = ₹15
NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2 – Quadratic Equations, has been designed by the NCERT to test the knowledge of the student on the topic – Solution of a Quadratic Equation by Factorisation
NCERT Solutions for Class 10 Maths
- NCERT Solutions Class 10 Maths Chapter 4 Ex. 4.1 Quadratic Equations
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- NCERT Solutions Class 10 Maths Chapter 4 Ex. 4.3 Quadratic Equations
- NCERT Solutions Class 10 Maths Chapter 4 Ex. 4.4 Quadratic Equations
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