Looking for accurate SEBA Class 10 Maths Chapter 4 Exercise 4.1 (Quadratic Equations) textual solutions for the 2026–27 academic year? Access step-by-step solutions for checking quadratic equations, framing mathematical word problems into standard quadratic form $ax^2 + bx + c = 0$, and detailed calculation notes designed to help you score top marks in your HSLC board exams.
EXERCISE 4.1
1. Check whether the following are quadratic equations :
(i) (x + 1)² = 2(x – 3)
LHS = (x + 1)² = x² + 2x + 1
RHS = 2(x – 3) = 2x – 6
x² + 2x + 1 = 2x – 6
x² + 7 = 0
It is of the form ax² + bx + c = 0, where a = 1 ≠ 0.
Ans: It is a quadratic equation.
(ii) x² – 2x = (-2)(3 – x)
LHS = x² – 2x
RHS = -6 + 2x
x² – 2x = -6 + 2x
x² – 4x + 6 = 0
It is of the form ax² + bx + c = 0, where a = 1 ≠ 0.
Ans: It is a quadratic equation.
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
LHS = x² + x – 2x – 2 = x² – x – 2
RHS = x² + 3x – x – 3 = x² + 2x – 3
x² – x – 2 = x² + 2x – 3
-3x + 1 = 0
It is a linear equation, not of the form ax² + bx + c = 0 (since a = 0).
Ans: It is not a quadratic equation.
(iv) (x – 3)(2x + 1) = x(x + 5)
LHS = 2x² + x – 6x – 3 = 2x² – 5x – 3
RHS = x² + 5x
2x² – 5x – 3 = x² + 5x
x² – 10x – 3 = 0
It is of the form ax² + bx + c = 0, where a = 1 ≠ 0.
Ans: It is a quadratic equation.
Class 10 Maths Chapter 4 Exercise 4.1 Question Answer SEBA 2026-27
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
LHS = 2x² – 6x – x + 3 = 2x² – 7x + 3
RHS = x² – x + 5x – 5 = x² + 4x – 5
2x² – 7x + 3 = x² + 4x – 5
x² – 11x + 8 = 0
It is of the form ax² + bx + c = 0, where a = 1 ≠ 0.
Ans: It is a quadratic equation.
(vi) x² + 3x + 1 = (x – 2)²
LHS = x² + 3x + 1
RHS = x² – 4x + 4
x² + 3x + 1 = x² – 4x + 4
7x – 3 = 0
It is not of the form ax² + bx + c = 0 (since a = 0).
Ans: It is not a quadratic equation.
(vii) (x + 2)³ = 2x(x² – 1)
LHS = x³ + 3(x²)(2) + 3(x)(2²) + 2³ = x³ + 6x² + 12x + 8
RHS = 2x³ – 2x
x³ + 6x² + 12x + 8 = 2x³ – 2x
-x³ + 6x² + 14x + 8 = 0
The degree of the equation is 3.
Ans: It is not a quadratic equation.
(viii) x³ – 4x² – x + 1 = (x – 2)³
LHS = x³ – 4x² – x + 1
RHS = x³ – 3(x²)(2) + 3(x)(2²) – 2³ = x³ – 6x² + 12x – 8
x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
2x² – 13x + 9 = 0
It is of the form ax² + bx + c = 0, where a = 2 ≠ 0.
Ans: It is a quadratic equation.
Quadratic Equations Class 10 SEBA Ex 4.1 Solutions
2. Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Let the breadth of the rectangular plot = x m.
Then, the length of the plot = (2x + 1) m.
Area of rectangle = Length * Breadth
528 = (2x + 1)x
2x² + x = 528
2x² + x – 528 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Let the two consecutive positive integers be x and (x + 1).
Product of integers = 306
x(x + 1) = 306
x² + x = 306
x² + x – 306 = 0
SEBA Class 10 Maths Chapter 4.1 Solutions
(iii) Ram’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Ram’s present age.
Let Ram’s present age = x years.
Mother’s present age = (x + 26) years.
After 3 years:
Ram’s age = (x + 3) years
Mother’s age = (x + 26 + 3) = (x + 29) years
Given product of their ages = 360:
(x + 3)(x + 29) = 360
x² + 29x + 3x + 87 = 360
x² + 32x + 87 – 360 = 0
x² + 32x – 273 = 0
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Let the uniform speed of the train = x km/h.
Total distance = 480 km.
Time taken at original speed (t1) = 480/x hours
Time taken at reduced speed (t2) = 480/(x – 8) hours
According to the question:
t2 – t1 = 3
480/(x – 8) – 480/x = 3
480 * [1/(x – 8) – 1/x] = 3
480 * [(x – (x – 8)) / (x(x – 8))] = 3
480 * [8 / (x² – 8x)] = 3
3840 / (x² – 8x) = 3
3(x² – 8x) = 3840
x² – 8x = 1280
x² – 8x – 1280 = 0
SEBA Class 10 Maths Chapter 4 Ex 4.1 Notes
3. For what value of p the equation (p – 2)x² + 3x + 5 = 0 can not be quadratic?
(a) 1
(b) 2
(c) – 2
(d) 0
For a standard quadratic equation ax² + bx + c = 0, the condition is a ≠ 0.
Here, a = (p – 2).
The equation will not be quadratic if a = 0:
p – 2 = 0 => p = 2
Ans: (b) 2
HSLC Class 10 Maths Chapter 4 Exercise 4.1 Answers
4. Which of the following are quadratic equations?
Choose the correct option.
(a) (i) and (iv)
(b) (i) and (ii)
(c) (i) and (iii)
(d) (ii) and (iv)
(i) (x + 1)² = 2(x – 4)
(i) (x + 1)² = 2(x – 4)
=> x² + 2x + 1 = 2x – 8
=> x² + 9 = 0 (Quadratic)
(ii) (x – 3)(x + 1) = (x + 2)(x – 3)
= (x + 2)(x – 3)
=> x² – 2x – 3 = x² – x – 6
=> -x + 3 = 0 (Linear)
(iii) (x – 2)² + 1 = 2x – 4
=> x² – 4x + 4 + 1 = 2x – 4
=> x² – 6x + 9 = 0 (Quadratic)
(iv) x(x + 3) + 7 = (x + 2)(x – 2)
=> x² + 3x + 7 = x² – 4
=> 3x + 11 = 0 (Linear)
Equations (i) and (iii) are quadratic.
Ans: (c) (i) and (iii)
📐 Updated Exercise 4.1 Solutions Notice: This page features complete, step-by-step SEBA Class 10 Maths Chapter 4 (Quadratic Equations) Exercise 4.1 Solutions updated for the 2026–27 academic session. All algebraic simplifications, standard form verifications, and word problem equations strictly follow the latest revised Assam Board (SEBA) textbook.
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