Looking for accurate SEBA Class 10 Maths Chapter 4 Exercise 4.2 (Quadratic Equations) textual solutions for the 2026–27 academic year? Access step-by-step solutions for finding roots by splitting the middle term, solving quadratic word problems, and mastering factorization techniques designed to help you score top marks in your HSLC board exams.
1. Find the roots of the following quadratic equations by factorisation:
(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
Roots: 5, -2
(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 = -2 or x = 3/2
Roots: -2, 3/2
(iii) √2x² + 7x + 5√2 = 0
√2 x² + 5x + 2x + 5√2 = 0
x(√2 x + 5) + √2(√2 x + 5) = 0
(√2 x + 5)(x + √2) = 0
√2 x + 5 = 0 or x + √2 = 0
x = -5/√2 or x = -√2
Roots: -5/√2, -√2
(iv) 2x² – x + 1/8 = 0
Multiplying the equation by 8:
16x² – 8x + 1 = 0
16x² – 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
Roots: 1/4, 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
x = 1/10 or x = 1/10
Roots: 1/10, 1/10
(vi) 2x² – 7x + 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 = 2 or x = 3/2
Roots: 2, 3/2
(vii) x² – 10x – 96 = 0
x² – 16x + 6x – 96 = 0
x(x – 16) + 6(x – 16) = 0
(x – 16)(x + 6) = 0
x – 16 = 0 or x + 6 = 0
x = 16 or x = -6
Roots: 16, -6
(viii) √3x² + 10x + 7√3 = 0
√3 x² + 3x + 7x + 7√3 = 0
√3 x(x + √3) + 7(x + √3) = 0
(x + √3)(√3 x + 7) = 0
x + √3 = 0 or √3 x + 7 = 0
x = -√3 or x = -7/√3
Roots: -√3, -7/√3
(ix) x² + 2√2x + 2 = 0
x² + √2 x + √2 x + 2 = 0
x(x + √2) + √2(x + √2) = 0
(x + √2)(x + √2) = 0
x + √2 = 0 or x + √2 = 0
x = -√2 or x = -√2
Roots: -√2, -√2
(x) 14x + 5 – 3x² = 0
3x² – 14x – 5 = 0
3x² – 15x + x – 5 = 0
3x(x – 5) + 1(x – 5) = 0
(x – 5)(3x + 1) = 0
x – 5 = 0 or 3x + 1 = 0
x = 5 or x = -1/3
Roots: 5, -1/3
SEBA Class 10 Maths Chapter 4.2 Solutions
2. Solve the problems given in Example 1.
3. Find two numbers whose sum is 27 and product is 182.
Let the first number be x.
Then, the second number = 27 – x.
According to the question:
x(27 – x) = 182
27x – x² = 182
x² – 27x + 182 = 0
x² – 14x – 13x + 182 = 0
x(x – 14) – 13(x – 14) = 0
(x – 14)(x – 13) = 0
x = 14 or x = 13
If x = 14, second number = 27 – 14 = 13.
If x = 13, second number = 27 – 13 = 14.
Answer: The two numbers are 13 and 14.
4. Find two consecutive positive integers, sum of whose squares is 365.
Let the two consecutive positive integers be x and x + 1.
According to the question:
x² + (x + 1)² = 365
x² + x² + 2x + 1 = 365
2x² + 2x + 1 – 365 = 0
2x² + 2x – 364 = 0
Dividing by 2:
x² + x – 182 = 0
x² + 14x – 13x – 182 = 0
x(x + 14) – 13(x + 14) = 0
(x + 14)(x – 13) = 0
x = -14 or x = 13
Since integers are positive, reject x = -14.
Therefore, x = 13.
Second integer = 13 + 1 = 14.
Answer: The required integers are 13 and 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.
Let base of the right triangle = x cm.
Altitude = (x – 7) cm.
Hypotenuse = 13 cm.
By Pythagoras theorem:
(Base)² + (Altitude)² = (Hypotenuse)²
x² + (x – 7)² = 13²
x² + x² – 14x + 49 = 169
2x² – 14x + 49 – 169 = 0
2x² – 14x – 120 = 0
Dividing 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 or x = -5
Length cannot be negative, so x = 12.
Base = 12 cm
Altitude = 12 – 7 = 5 cm
Answer: The other two sides are 12 cm and 5 cm.
Class 10 Maths Chapter 4 Exercise 4.2 Question Answer SEBA 2026-27
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 number of articles produced on that day = x.
Cost of production of each article = Rs (2x + 3).
Total cost of production = Rs 90.
According to the question:
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
x = -15/2 or x = 6
Since the number of articles cannot be negative, x = 6.
Cost of each article = 2(6) + 3 = 15.
Answer: Number of articles = 6, Cost of each article = Rs 15.
Choose the correct option :
7. If x² – 2px + p² = 0 then the value of p/x is
(a) 0
(b) -1
(c) 1
(d) 2
Given equation:
x² – 2px + p² = 0
(x – p)² = 0
x – p = 0 => x = p
Now, p/x = p/p = 1
Answer: (c) 1
Quadratic Equations Class 10 SEBA Ex 4.2 Solutions
8. A student has done through the following steps to find the roots of the equation x² – 3x – 10 = 0 by factorization method
Step 1 : x² – 3x – 10 = 0
Step 2 : x² – 5x – 2x – 10 = 0
Step 3 : x(x – 5) – 2(x – 5) = 0
Step 4 : (x – 5)(x – 2) = 0
Step 5 : x = 5 and x = 2
In which step did the student make the first mistake?
(a) step : 2
(b) step : 3
(d) step : 4
(d) step : 5
Step 2 states: x² – 5x – 2x – 10 = 0
Splitting -3x into -5x – 2x gives -7x, which is incorrect (it should be -5x + 2x).
Answer: (a) step : 2
9. Find the sum and product of the roots of the quadratic equation
2x² – 9x + 4 = 0
Given equation:
2x² – 9x + 4 = 0
Here a = 2, b = -9, c = 4.
Sum of roots = -b/a = -(-9)/2 = 9/2
Product of roots = c/a = 4/2 = 2
Answer: Sum of roots = 9/2, Product of roots = 2
SEBA Class 10 Maths Chapter 4 Ex 4.2 Notes
10. If one root of the quadratic equation 2x² + kx – 6 = 0 is 2 then find the value of k. Also find the other root.
- Given equation: 2x² + kx – 6 = 0
Since x = 2 is a root, substituting x = 2:
2(2)² + k(2) – 6 = 0
2(4) + 2k – 6 = 0
8 + 2k – 6 = 0
2k + 2 = 0
2k = -2 => k = -1
Substitute k = -1 back into the equation:
2x² – x – 6 = 0
2x² – 4x + 3x – 6 = 0
2x(x – 2) + 3(x – 2) = 0
(x – 2)(2x + 3) = 0
x = 2 or x = -3/2
Answer: k = -1, Other root = -3/2
HSLC Class 10 Maths Chapter 4 Exercise 4.2 Answers
📐 Updated Exercise 4.2 Solutions Notice: This page features complete, step-by-step SEBA Class 10 Maths Chapter 4 (Quadratic Equations) Exercise 4.2 Solutions updated for the 2026–27 academic session. All middle-term splitting methods, root factorizations, and word problem steps strictly follow the latest revised Assam Board (SEBA) textbook.
💡 Ace Your HSLC Class 10 Maths Exam:
- Have a doubt about splitting the middle term, handling square root factors, or setting up equations in Exercise 4.2? Drop your questions in the comments section below!
- Save and bookmark this solution guide for fast revision before your unit tests, half-yearly, pre-board, and HSLC final examinations.
- Share this link with your classmates and WhatsApp study groups to help them master Class 10 Mathematics!