NCERT Class 10 Maths – Exercise 4.1 Solutions

NCERT Class 10 Maths

Chapter 4 – Quadratic Equations | Exercise 4.1

(Rationalized Syllabus 2025-26)

Q1

Check whether the following are quadratic equations:

(i) $(x + 1)^2 = 2(x – 3)$
Expand using $(a+b)^2$:
$$x^2 + 2x + 1 = 2x – 6$$
Simplifying:
$$x^2 + 2x – 2x + 1 + 6 = 0$$
$$x^2 + 7 = 0$$
It is of the form $ax^2 + bx + c = 0$ (where $a \neq 0$).
✔️ Yes, it is a quadratic equation
(ii) $x^2 – 2x = (-2)(3 – x)$
$$x^2 – 2x = -6 + 2x$$
Bringing terms to LHS:
$$x^2 – 2x – 2x + 6 = 0$$
$$x^2 – 4x + 6 = 0$$
It is of the form $ax^2 + bx + c = 0$.
✔️ Yes, it is a quadratic equation
(iii) $(x – 2)(x + 1) = (x – 1)(x + 3)$
Expand both sides:
$$x^2 + x – 2x – 2 = x^2 + 3x – x – 3$$
$$x^2 – x – 2 = x^2 + 2x – 3$$
Cancel $x^2$ from both sides:
$$-x – 2 = 2x – 3 \Rightarrow 3x – 1 = 0$$
This is a linear equation (degree 1), not quadratic.
✖️ No, it is not a quadratic equation
(iv) $(x – 3)(2x + 1) = x(x + 5)$
Expand terms:
$$2x^2 + x – 6x – 3 = x^2 + 5x$$
$$2x^2 – 5x – 3 = x^2 + 5x$$
Rearranging:
$$x^2 – 10x – 3 = 0$$
✔️ Yes, it is a quadratic equation
(v) $(2x – 1)(x – 3) = (x + 5)(x – 1)$
Expand both sides:
$$2x^2 – 6x – x + 3 = x^2 – x + 5x – 5$$
$$2x^2 – 7x + 3 = x^2 + 4x – 5$$
Rearranging:
$$x^2 – 11x + 8 = 0$$
✔️ Yes, it is a quadratic equation
(vi) $x^2 + 3x + 1 = (x – 2)^2$
Expand RHS:
$$x^2 + 3x + 1 = x^2 – 4x + 4$$
Cancel $x^2$:
$$3x + 1 = -4x + 4$$
$$7x – 3 = 0$$
✖️ No, it is not a quadratic equation
(vii) $(x + 2)^3 = 2x(x^2 – 1)$
Expand using $(a+b)^3$:
$$x^3 + 8 + 6x(x+2) = 2x^3 – 2x$$
$$x^3 + 8 + 6x^2 + 12x = 2x^3 – 2x$$
Rearranging:
$$-x^3 + 6x^2 + 14x + 8 = 0$$
Degree is 3 (Cubic), not 2.
✖️ No, it is not a quadratic equation
(viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$
Expand RHS:
$$x^3 – 4x^2 – x + 1 = x^3 – 8 – 6x(x-2)$$
$$x^3 – 4x^2 – x + 1 = x^3 – 8 – 6x^2 + 12x$$
Cancel $x^3$ and rearrange:
$$2x^2 – 13x + 9 = 0$$
✔️ Yes, it is a quadratic equation
Q2

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 be $x$ meters.
Then length $= (2x + 1)$ meters.
Area = Length × Breadth = 528
$$(2x + 1) \cdot x = 528$$
$$2x^2 + x – 528 = 0$$
Equation: $2x^2 + x – 528 = 0$
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Let the first integer be $x$.
The next consecutive integer is $(x + 1)$.
$$x(x + 1) = 306$$
$$x^2 + x – 306 = 0$$
Equation: $x^2 + x – 306 = 0$
(iii) Rohan’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 Rohan’s present age.
Let Rohan’s present age be $x$ years.
Mother’s present age $= (x + 26)$ years.
After 3 years:
Rohan’s age $= x + 3$
Mother’s age $= (x + 26) + 3 = x + 29$
Product is 360:
$$(x + 3)(x + 29) = 360$$
$$x^2 + 29x + 3x + 87 = 360$$
$$x^2 + 32x – 273 = 0$$
Equation: $x^2 + 32x – 273 = 0$
(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 be $x$ km/h.
Time taken = $\frac{480}{x}$ hours.
Reduced speed $= (x – 8)$ km/h.
New time = $\frac{480}{x – 8}$ hours.
According to the condition (New time – Old time = 3):
$$\frac{480}{x – 8} – \frac{480}{x} = 3$$
Simplifying:
$$480 \left( \frac{x – (x – 8)}{x(x – 8)} \right) = 3$$
$$480 \cdot 8 = 3x(x – 8)$$
$$3840 = 3x^2 – 24x$$
$$3x^2 – 24x – 3840 = 0$$
Divide by 3:
$$x^2 – 8x – 1280 = 0$$
Equation: $x^2 – 8x – 1280 = 0$
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