NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2
Access the comprehensive NCERT Solutions for Class 10 Mathematics Chapter 4, Exercise 4.2 on Quadratic Equations. These solutions are offered free of charge and prove highly beneficial in completing homework assignments. The experienced instructors at School Notes Wallah have meticulously prepared the Class 10 Maths NCERT Solutions for Exercise 4.2. You can find detailed explanations for all the questions in Chapter 4 of the NCERT textbook on Quadratic Equations.
Topics and Sub Topics in Class 10 Maths Chapter 4 Quadratic Equations:
| Section Name | Topic Name |
| 4 | Quadratic Equations |
| 4.1 | Introduction |
| 4.2 | Quadratic Equations |
| 4.3 | Solution of a Quadratic Equation by Factorisation |
| 4.4 | Solution of a Quadratic Equation by Completing the Square |
| 4.5 | Nature of Roots |
| 4.6 | Summary |
You can also download the free PDF of Chapter 4 Ex 4.2 Quadratic Equations NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 4 |
| Chapter Name | Quadratic Equations |
| Exercise | Ex 4.2 |
| Number of Questions Solved | 6 |
| Category | NCERT Solutions |
NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2.
EXERCISE 4.2
Q.1. Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) 
(iv) 2x2 – x + 8 = 0
(v) 100x2 – 20x + 1 = 0
Sol. (i) x2 – 3x – 10 = 0
We have:
x2 – 3x – 10 = 0
⇒ x2 – 5x + 2x – 10 = 0
⇒ x (x – 5) + 2(x – 5) = 0
⇒ (x – 5) (x + 2) = 0
Either x – 5 = 0 ⇒ x = 5
or x + 2 = 0 ⇒ x = –2
Thus, the required roots are x = 5 and x = –2.
(ii) 2x2 + x – 6 = 0
We have:
2x2 + x – 6 = 0
2x2 + 4x – 3x – 6 = 0
2x(x + 2) – 3 (x + 2) = 0
(x + 2) (2x – 3) = 0
Either x + 2 = 0 ⇒ x = –2
or 2x – 3 = 0 ⇒ x = –2
Thus, the required roots are x = –2 and 
(v) 100x2 – 20x + 1 = 0
We have:
100x2 – 20x + 1 = 0
100x2 – 10x – 10x + 1 = 0
10x(10x – 1) – 1 (10x – 1) = 0
(10x – 1) (10x – 1) = 0
(10x – 1) = 0 and (10x – 1) = 0
Q.2. Solve the problems given in Example 1.
Sol. (i) We have:
x2 – 45x + 324 = 0
x2 – 9x – 36x + 324 = 0
x(x – 9) – 36(x – 9) = 0
(x – 9) (x – 36) = 0
Either x – 9 = 0 ⇒ x = 9
or x – 36 = 0 ⇒ x = 36
Thus, x = 9 and × = 36
(ii) We have:
x2 – 55x + 750 = 0
x2 – 30x – 25x + 750 = 0 ∵(– 30) × (– 25) = 750 and
x (x – 30) – 25 (x – 30) = 0 (– 30) + (– 25) = – 55
(x – 30) (x – 25) = 0
Either x – 30 = 0 ⇒ x = 30
or x – 25 = 0 ⇒ x = 25
Thus, x = 30 and x = 25.
Q.3. Find two numbers whose sum is 27 and product is 182.
Sol. Here, sum of the numbers is 27.
Let one of the numbers be x.
∴ Other number = 27 – x
According to the condition,
Product of the numbers = 182
⇒ x (27 – x) = 182
⇒ 27x – x2 = 182
⇒ –x2 + 27x – 182 = 0
⇒ x2 – 27x + 182 = 0
⇒ x2 – 13x – 14x + 182 = 0 –27 = (–13) + (– 14) and
⇒ x (x – 13) – 14 (x – 13) = 0 (– 13) × (– 14) = 182
⇒ (x – 13) (x – 14) = 0
Either x – 13 = 0 ⇒ x = 13
or x – 14 = 0 ⇒ x = 14
Thus, the required numbers are 13 and 14.
Q.4 Find two consecutive positive integers, sum of whose squares is 365.
Sol. Let the two consecutive positive integers be x and (x + 1).
Since the sum of the squares of the numbers = 365
x2 + (x + 1)2 = 365
x2 + [x2 + 2x + 1] = 365
x2 + x2 + 2x + 1 = 365
2x2 + a + 1 – 365 = 0
2x2 + 2x – 364 =0
x2 + x – 182 = 0
x2 + 14x – 13x – 182 = 0 ∵ +14 –13 = 1 and 14 × (–13) = ‐ 182
x(x + 14) –13 (x + 14) = 0
(x +14) (x – 13) = 0
Either x + 14 = 0 ⇒ x = – 14
or x – 13 = 0 = x = 13
Since x has to be a positive integer
∴ x = 13
⇒ x + 1 = 13 + 1 =1 4
Thus, the required consecutive positive integers are 13 and 14.
Q.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.
Sol. Let the base of the given right triangle be ‘x’ cm.
∴ Its height = (x – 7) cm
Squaring both sides, we get
169 = x2 + (x – 7)2
⇒ 169 = x2 + x2 – 14x + 49
⇒ 14x + 49 – 169 = 0
⇒ 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
Either x –12 = 0 ⇒ x=12
or x + 5 = 0 ⇒ x = –5
But the side of triangle can never be negative,
⇒ x = 12
∴ Length of the base = 12 cm
⇒ Length of the height = (12 – 7) cm = 5 cm
Thus, the required base = 12 cm and height = 5 cm,
Q.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.
Sol. Let the number of articles produced in a day = x
∴ Cost of production of each article = Z (2x + 3)
According to the condition,
Total cost = 90
⇒ 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
⇒ x – 6 = 0 ⇒ x = 6
But the number of articles cannot be negative.
⇒ x = 6
∴ Cost of each article = Rs (2 × 6 + 3) = Rs. 15
Thus, the required number of articles produced is 6 and the cost of each article is Rs. 15.
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