SEBA Class 10 Maths Chapter 2 Exercise 4.1 Solutions (2026–27) – Updated HSLC Answers PDF & Important Questions

Q1(i). Check whether the following is a quadratic equation:
\((x + 1)^2 = 2(x - 3)\)

Answer:
Given equation:
\[ (x + 1)^2 = 2(x - 3) \]
Expand both sides:
\[ x^2 + 2x + 1 = 2x - 6 \]
Bring all terms to one side:
\[ x^2 + 2x + 1 - 2x + 6 = 0 \] \[ x^2 + 7 = 0 \]
This is of the form:
\[ ax^2 + bx + c = 0 \] where \(a = 1,\; b = 0,\; c = 7\) and \(a \ne 0\)

Therefore, it is a quadratic equation.

Q1(i). Check whether the following is a quadratic equation:
\((x + 1)^2 = 2(x - 3)\)

Answer:
Given equation:
\[ (x + 1)^2 = 2(x - 3) \]
Expand both sides:
\[ x^2 + 2x + 1 = 2x - 6 \]
Bring all terms to one side:
\[ x^2 + 2x + 1 - 2x + 6 = 0 \] \[ x^2 + 7 = 0 \]
This is of the form:
\[ ax^2 + bx + c = 0 \] where \(a = 1,\; b = 0,\; c = 7\) and \(a \ne 0\)

Therefore, 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 is one more than twice its breadth. Find the quadratic equation.

Answer:
Let breadth of the rectangular plot be \(x\) m.
Then its length will be = \(2x + 1\) m.
we know, Area = length × breadth
\[ x(2x + 1) = 528 \] \[ 2x^2 + x = 528 \] \[ 2x^2 + x - 528 = 0 \]
Therefore, required quadratic equation is \[ 2x^2 + x - 528 = 0 \]

(ii) The product of two consecutive positive integers is 306. Find the quadratic equation.

Answer:
Let the two consecutive positive integer be \(x\) and \[ x(x + 1) = 306 \] It is given that the product of these number are 306 \[ Therefore, x^2 + x = 306 \] \[ x^2 + x - 306 = 0 \]
Therefore, required quadratic equation is \[ x^2 + x - 306 = 0 \]

(iii) Ram’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. Find the quadratic equation.

Answer:
Let Ram’s present age be \(x\) years.
then his mother’s age = \(x + 26\).
After 3 years:
Ram = \(x + 3\)
Mother = \(x + 29\)
\[ (x + 3)(x + 29) = 360 \] \[ x^2 + 32x + 87 = 360 \] \[ x^2 + 32x - 273 = 0 \]
Therefore, required quadratic equation is \[ x^2 + 32x - 273 = 0 \]

(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. Find the quadratic equation.

Answer:
Let speed of train be \(x\) km/h.
we know, Time = Distance / Speed
\[ \text{Time} = \frac{480}{x} \] Now new speed = \(x - 8\)
and new time = \( \frac{480}{x - 8} \)
Given:
\[ \frac{480}{x - 8} = \frac{480}{x} + 3 \]
Multiply by \(x(x - 8)\):
\[ 480x = 480(x - 8) + 3x(x - 8) \] \[ 480x = 480x - 3840 + 3x^2 - 24x \] \[ 0 = 3x^2 - 24x - 3840 \] Divide by 3:
\[ x^2 - 8x - 1280 = 0 \]
Therefore, required quadratic equation is \[ x^2 - 8x - 1280 = 0 \]

Q3. For what value of p, the equation \((p - 2)x^2 + 3x + 5 = 0\) cannot be quadratic?

(a) 1
(b) 2
(c) -2
(d) 0

Answer: (b) 2

Q4. Which of the following are quadratic equations?

(i) \((x + 1)^2 = 2(x - 4)\)
(ii) \((x - 3)(x + 1) = (x + 2)(x - 3)\)
(iii) \((x - 2)^2 + 1 = 2x - 4\)
(iv) \(x(x + 3) + 7 = (x + 2)(x - 2)\)

(a) (i) and (iv)
(b) (i) and (ii)
(c) (i) and (iii)
(d) (ii) and (iv)

Answer: (c) (i) and (iii)