Class 10 Maths Chapter 4 Exercise 4.1 Solutions English Medium | Quadratic Equations SEBA 2026–2027
Quadratic Equations form one of the most important chapters in Class 10 Mathematics under the SCERT Assam / SEBA 2026–2027 syllabus. This chapter builds a strong foundation for higher mathematics and competitive exams.
✨ What You Will Learn in Exercise 4.1
Exercise 4.1 focuses on:
Identifying quadratic equations
Writing equations in standard form
Finding coefficients
Checking whether an equation is quadratic or not
📚 Exercise 4.1 – Complete Solutions in English Medium
(i) \((x + 1)^2 = 2(x - 3)\)
(ii) \(x^2 - 2x = (-2)(3 - x)\)
(iii) \((x - 2)(x + 1) = (x - 1)(x + 3)\)
(iv) \((x - 3)(2x + 1) = x(x + 5)\)
(v) \((2x - 1)(x - 3) = (x + 5)(x - 1)\)
(vi) \(x^2 + 3x + 1 = (x - 2)^2\)
(vii) \((x + 2)^3 = 2x(x^2 - 1)\)
(viii) \(x^3 - 4x^2 - x + 1 = (x - 2)^3\)
Solutions:
(i) Given, \((x + 1)^2 = 2(x - 3)\)
Using the formula \((a+b)^2 = a^2 + 2ab + b^2\):
\(\Rightarrow x^2 + 2x + 1 = 2x - 6\)
\(\Rightarrow x^2 + 7 = 0\)
The above equation is in the form of \(ax^2 + bx + c = 0\).
Therefore, the given equation is a quadratic equation.
(ii) Given, \(x^2 - 2x = (-2)(3 - x)\)
\(\Rightarrow x^2 - 2x = -6 + 2x\)
\(\Rightarrow x^2 - 4x + 6 = 0\)
The above equation is in the form of \(ax^2 + bx + c = 0\).
Therefore, the given equation is a quadratic equation.
(iii) Given, \((x - 2)(x + 1) = (x - 1)(x + 3)\)
By multiplication:
\(\Rightarrow x^2 + x - 2x - 2 = x^2 + 3x - x - 3\)
\(\Rightarrow x^2 - x - 2 = x^2 + 2x - 3\)
\(\Rightarrow -3x + 1 = 0\) or \(3x - 1 = 0\)
The above equation is not in the form of \(ax^2 + bx + c = 0\).
Therefore, the given equation is not a quadratic equation.
(iv) Given, \((x - 3)(2x + 1) = x(x + 5)\)
\(\Rightarrow 2x^2 + x - 6x - 3 = x^2 + 5x\)
\(\Rightarrow 2x^2 - 5x - 3 = x^2 + 5x\)
\(\Rightarrow x^2 - 10x - 3 = 0\)
The above equation is in the form of \(ax^2 + bx + c = 0\).
Therefore, the given equation is a quadratic equation.
(v) Given, \((2x - 1)(x - 3) = (x + 5)(x - 1)\)
\(\Rightarrow 2x^2 - 6x - x + 3 = x^2 - x + 5x - 5\)
\(\Rightarrow 2x^2 - 7x + 3 = x^2 + 4x - 5\)
\(\Rightarrow x^2 - 11x + 8 = 0\)
The above equation is in the form of \(ax^2 + bx + c = 0\).
Therefore, the given equation is a quadratic equation.
(vi) Given, \(x^2 + 3x + 1 = (x - 2)^2\)
Using the formula \((a-b)^2 = a^2 - 2ab + b^2\):
\(\Rightarrow x^2 + 3x + 1 = x^2 - 4x + 4\)
\(\Rightarrow 7x - 3 = 0\)
The above equation is not in the form of \(ax^2 + bx + c = 0\).
Therefore, the given equation is not a quadratic equation.
(vii) Given, \((x + 2)^3 = 2x(x^2 - 1)\)
Using the formula \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\):
\(\Rightarrow x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 = 2x^3 - 2x\)
\(\Rightarrow x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x\)
\(\Rightarrow -x^3 + 6x^2 + 14x + 8 = 0\)
Since the highest power (degree) of the equation is 3, it is not in the form of \(ax^2 + bx + c = 0\).
Therefore, it is not a quadratic equation.
(viii) Given, \(x^3 - 4x^2 - x + 1 = (x - 2)^3\)
Using the formula \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\):
\(\Rightarrow x^3 - 4x^2 - x + 1 = x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3\)
\(\Rightarrow x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8\)
\(\Rightarrow 2x^2 - 13x + 9 = 0\)
The above equation is in the form of \(ax^2 + bx + c = 0\).
Therefore, the given equation is a quadratic equation.
Solutions:
(i) The area of a rectangular plot is 528 m³. 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.
As we know, Area of rectangle = Length \(\times\) Breadth = 528 m².
According to the problem,
\((2x + 1) \times x = 528\)
\(\Rightarrow 2x^2 + x = 528\)
\(\Rightarrow 2x^2 + x - 528 = 0\)
This is the required quadratic equation.
Solutions:
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Let the first integer = \(x\).Then, the next consecutive positive integer = \(x + 1\).
Product of the two integers = \(x \times (x + 1) = 306\).
\(\Rightarrow x^2 + x = 306\)
\(\Rightarrow x^2 + x - 306 = 0\)
This is the required quadratic equation.
Solutions:
(iii) Ram's mother is 26 years older than him. The product of their ages (in yea 3 years from now will be 360. We would like to find Ram's present age.
Let Rohan's present age = \(x\) years.Then, Rohan’s mother’s present age = \(x + 26\) years.
After 3 years:
Rohan’s age = \(x + 3\) years.
Mother’s age = \(x + 26 + 3 = x + 29\) years.
According to the problem,
\((x + 3)(x + 29) = 360\)
\(\Rightarrow x^2 + 29x + 3x + 87 = 360\)
\(\Rightarrow x^2 + 32x + 87 - 360 = 0\)
\(\Rightarrow x^2 + 32x - 273 = 0\)
This is the required quadratic equation.
Solutions:
(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 speed of the train = \(x\) km/h.Time taken to travel 480 km = \(\frac{480}{x}\) hours.
If the speed is 8 km/h less, the new speed = \((x - 8)\) km/h.
According to the second condition, the train takes 3 hours more. So, time = \(\left(\frac{480}{x} + 3\right)\) hours.
As we know, Speed \(\times\) Time = Distance:
\((x - 8) \left(\frac{480}{x} + 3\right) = 480\)
\(\Rightarrow 480 + 3x - \frac{3840}{x} - 24 = 480\)
\(\Rightarrow 3x - \frac{3840}{x} - 24 = 0\)
Multiplying the entire equation by \(x\):
\(\Rightarrow 3x^2 - 3840 - 24x = 0\)
\(\Rightarrow 3x^2 - 24x - 3840 = 0\)
Dividing by 3:
\(\Rightarrow x^2 - 8x - 1280 = 0\)
This is the required quadratic equation.
(a) \(1\)
(b) \(2\)
(c) \(-2\)
(d) \(0\)
Solution:
The given equation is:
\((p - 2)x^2 + 3x + 5 = 0\)
We know that a general equation of the form \(ax^2 + bx + c = 0\) is called a quadratic equation only if the coefficient of \(x^2\) is not zero (i.e., \(a \neq 0\)).
For the given equation to not be a quadratic equation, the coefficient of \(x^2\) must be equal to zero. Therefore:
\(a = 0\)
\(\Rightarrow p - 2 = 0\)
\(\Rightarrow p = 2\)
Thus, for \(p = 2\), the equation will not be a quadratic equation because the \(x^2\) term will vanish.
Correct Option: (b) \(2\)
(i) \((x + 1)^2 = 2(x - 4)\)
(ii) \((x - 3)(x + 1) = (x + 2)(x - 3)\)
(iii) \((x - 2)^2 + 1 = 2x - 4\)
(iv) \(x(x + 3) + 7 = (x + 2)(x - 2)\)
Choose the correct option:
(a) (i) and (iv)
(b) (i) and (ii)
(c) (i) and (iii)
(d) (ii) and (iv)
Solution:
To identify a quadratic equation, we simplify each expression to see if it fits the form \(ax^2 + bx + c = 0\) (where \(a \neq 0\)).
(i) \((x + 1)^2 = 2(x - 4)\)
\(\Rightarrow x^2 + 2x + 1 = 2x - 8\)
\(\Rightarrow x^2 + 9 = 0\)
Since it has an \(x^2\) term, it is a Quadratic Equation.
(ii) \((x - 3)(x + 1) = (x + 2)(x - 3)\)
\(\Rightarrow x^2 + x - 3x - 3 = x^2 - 3x + 2x - 6\)
\(\Rightarrow x^2 - 2x - 3 = x^2 - x - 6\)
\(\Rightarrow -x + 3 = 0\)
The \(x^2\) terms cancel out, so it is not a Quadratic Equation.
(iii) \((x - 2)^2 + 1 = 2x - 4\)
\(\Rightarrow x^2 - 4x + 4 + 1 = 2x - 4\)
\(\Rightarrow x^2 - 6x + 9 = 0\)
It has an \(x^2\) term, so it is a Quadratic Equation.
(iv) \(x(x + 3) + 7 = (x + 2)(x - 2)\)
\(\Rightarrow x^2 + 3x + 7 = x^2 - 4\)
\(\Rightarrow 3x + 11 = 0\)
The \(x^2\) terms cancel out, so it is not a Quadratic Equation.
Thus, only (i) and (iii) are quadratic equations.
Correct Option: (c) (i) and (iii)
Class 10 Maths Chapter 4 – Quadratic Equations (Exercise 4.1) FAQs
Exercise 4.1 mainly focuses on identifying quadratic equations and converting given equations into the standard form (ax² + bx + c = 0).
A quadratic equation is an equation of the form ax² + bx + c = 0, where a ≠ 0. The presence of x² makes it a second-degree equation.
Yes, it is very important. Usually, 1–2 mark questions are asked from this exercise, and it is easy to score if concepts are clear.
If an equation contains x² and can be written in the form ax² + bx + c = 0, then it is a quadratic equation.
Questions usually ask you to convert equations into standard form and check whether they are quadratic or not.
Practice regularly, understand each step clearly, revise formulas, and solve previous year questions.
You can find full step-by-step solutions on Digital Pipal Academy in a simple and easy-to-understand format.
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