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CBSE Important Questions Class 10 Maths Chapter 4 – Quadratic Equations

Class 10 Maths Chapter 4 Quadratic Equations is part of the CBSE Class 10 Maths syllabus. It explains what a quadratic equation is and the different ways of solving one: by factorisation (splitting the middle term) and by the quadratic formula. Students also learn about the discriminant D = b² − 4ac, which tells us the nature of the roots without actually solving the equation.

These CBSE important questions for Class 10 Maths Chapter 4 cover the nature of the roots, finding the sides of a right triangle using the Pythagoras theorem and a quadratic equation, factorising to find the roots, finding the value of k for equal roots, and the condition for real and equal roots using the discriminant. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

CBSE Important Questions Class 10 Maths Chapter 4 – Quadratic Equations

Q1. Given the quadratic equation (x − a)(x − b) + (x − b)(x − c) + (x − c)(x − a) = 0 and the condition that a = b = c, find the roots of the equation and determine their nature.

Answer: Putting a = b = c in the equation, each bracket becomes (x − a):
(x − a)(x − a) + (x − a)(x − a) + (x − a)(x − a) = 0
3(x − a)² = 0
(x − a)² = 0
So x = a and x = a.
The roots are x = a and x = a. Since both roots are the same, their nature is real and equal.

Q2. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, what are the measures of the other two sides?

Answer: Let the base be x cm. Then the altitude is (x − 7) cm.
By the Pythagoras theorem:
x² + (x − 7)² = 13²
x² + x² − 14x + 49 = 169
2x² − 14x − 120 = 0
Dividing by 2: x² − 7x − 60 = 0
Splitting the middle term: x² − 12x + 5x − 60 = 0
x(x − 12) + 5(x − 12) = 0
(x − 12)(x + 5) = 0
So x = 12 or x = −5. A length cannot be negative, so x = 12.
Base = 12 cm and altitude = 12 − 7 = 5 cm.

Q3. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, then the other two sides are
(a) 12.0 cm and 5.0 cm
(b) 24.0 cm and 25.0 cm
(c) 12.0 cm and 25.0 cm
(d) 24.0 cm and 5.0 cm

Answer: (a) 12.0 cm and 5.0 cm
As worked out above, solving x² − 7x − 60 = 0 gives the base as 12 cm and the altitude as 5 cm. We can check: 12² + 5² = 144 + 25 = 169 = 13² ✓

Q4. One side of a right triangle exceeds the other side by four inches. The hypotenuse is 20 inches. The length of the shorter side of the right triangle will be
(a) 8 inches
(b) 12 inches
(c) 16 inches
(d) 18 inches

Answer: (b) 12 inches
Let the shorter side be x. Then the other side is (x + 4).
x² + (x + 4)² = 20²
x² + x² + 8x + 16 = 400
2x² + 8x − 384 = 0
Dividing by 2: x² + 4x − 192 = 0
(x + 16)(x − 12) = 0
Since a length cannot be negative, x = 12 inches.

Q5. If f(x) = ax² + bx + c and g(x) = −ax² + bx + c where ac ≠ 0, then f(x) · g(x) = 0 has
(a) at least three real roots
(b) no real roots
(c) at least two real roots
(d) two real roots and two imaginary roots

Answer: (c) at least two real roots
The discriminant of f(x) is b² − 4ac and that of g(x) is b² + 4ac.
If ac is positive, then b² + 4ac is positive, so g(x) has real roots.
If ac is negative, then b² − 4ac is positive, so f(x) has real roots.
In either case at least one of the two polynomials has two real roots, so the product equation has at least two real roots.

Q6. The roots of the equation x² − 4ax + 4a² − b² = 0 are
(a) (a + b) and (a − b)
(b) (a + 2b) and (a − 2b)
(c) (2a + b) and (2a − b)
(d) (2a + 2b) and (2a − 2b)

Answer: (c) (2a + b) and (2a − b)
Group the terms to use the identity for the difference of two squares:
(x² − 4ax + 4a²) − b² = 0
(x − 2a)² − b² = 0
(x − 2a + b)(x − 2a − b) = 0
So x = 2a − b or x = 2a + b.

Q7. The values of k for which the equation 2x² − kx + x + 8 = 0 will have real and equal roots are
(a) 9 and −7
(b) only 9
(c) only −7
(d) −9 and 7

Answer: (a) 9 and −7
First write the equation in standard form:
2x² + (1 − k)x + 8 = 0
Here a = 2, b = (1 − k), c = 8.
For real and equal roots, the discriminant must be zero:
b² − 4ac = 0
(1 − k)² − 4(2)(8) = 0
(1 − k)² = 64
1 − k = ±8
If 1 − k = 8, then k = −7. If 1 − k = −8, then k = 9.
So k = 9 or k = −7.

Q8. Find the values of k for the quadratic equation kx(x − 2) + 6 = 0 having two equal roots.

Answer: Expand the equation first:
kx² − 2kx + 6 = 0
Here a = k, b = −2k, c = 6.
For two equal roots, the discriminant must be zero:
b² − 4ac = 0
(−2k)² − 4(k)(6) = 0
4k² − 24k = 0
4k(k − 6) = 0
So k = 0 or k = 6.
But if k = 0 the equation is no longer quadratic. So k = 6.

Q9. What is the discriminant of a quadratic equation? How does it tell us the nature of the roots?

Answer: For the quadratic equation ax² + bx + c = 0 (a ≠ 0), the discriminant is D = b² − 4ac.
• If D > 0, the roots are real and distinct (two different roots).
• If D = 0, the roots are real and equal (both roots are the same).
• If D < 0, the equation has no real roots (the roots are imaginary).
This lets us find the nature of the roots without actually solving the equation.

Q10. Solve the quadratic equation 2x² − 5x + 3 = 0 by factorisation.

Answer: We split the middle term. We need two numbers whose product is 2 × 3 = 6 and whose sum is −5. These are −2 and −3.
2x² − 2x − 3x + 3 = 0
2x(x − 1) − 3(x − 1) = 0
(x − 1)(2x − 3) = 0
So x − 1 = 0 or 2x − 3 = 0.
The roots are x = 1 and x = 3/2.

Q11. Write the quadratic formula. Use it to solve 3x² − 5x + 2 = 0.

Answer: Quadratic formula: For ax² + bx + c = 0, the roots are
x = [−b ± √(b² − 4ac)] / 2a
For 3x² − 5x + 2 = 0: a = 3, b = −5, c = 2.
D = b² − 4ac = 25 − 24 = 1
x = [5 ± √1] / 6
x = (5 + 1)/6 = 1 or x = (5 − 1)/6 = 2/3
So the roots are x = 1 and x = 2/3.

Q12. The sum of the squares of two consecutive positive integers is 365. Find the integers.

Answer: 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 − 364 = 0
Dividing by 2: x² + x − 182 = 0
Splitting the middle term: x² + 14x − 13x − 182 = 0
x(x + 14) − 13(x + 14) = 0
(x + 14)(x − 13) = 0
So x = −14 or x = 13. Since the integers are positive, x = 13.
The two integers are 13 and 14. (Check: 169 + 196 = 365 ✓)

Related Study Material on Extramarks

Q.1 What are the roots of quadratic equation 2x² + 5x + 2 = 0

(a) 2, (1/2)

(b) 2, (1/2)

(c) 2, (1/2)

(d) 2, (1/2)

Marks:1
Ans

2, (1/2)

Q.2 The area of a right angled triangle is 600 cm². If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Marks:4
Ans

Let the altitude of the right-angled triangle be x cm, then
Base = (x + 10) cm.
Area = 1/2 — base — height

$\begin{array}{l} Area= \frac{1}{2} (x + 10) . x {cm}^{2} \\ \frac{1}{2} x (x + 10) = 600 \\ x^{2} + 10 x = 1200 \\ x^{2} + 10 x 1200 = 0 \\ x^{2} + 40 x 30 x 1200 = 0 \\ x (x + 40) 30 (x + 40) = 0 \\ (x + 40) (x 30) = 0 \\ x = 30, 40 \\ Length cannot be negative \\ Altitude = x = 30 cm \\ Base = x + 10 = 40 cm \\ Hypotenuse = \sqrt{30^{2} + 40^{2}} = \sqrt{2500} = 50 cm \end{array}$

Q.3 Using quadratic formula solve the equation for x.
abx² + (b² – ac)x – bc = 0

Marks:2
Ans

$\begin{array}{l} Here, {abx}^{2} + (b^{2} ac) x bc = 0 \\ x = \frac{B \pm \sqrt{(B^{2} 4 AC)}}{2 A} \\ x = \frac{(b^{2} ac) \pm \sqrt{{(b^{2} ac)}^{2} 4 ab (bc)}}{2 ab} \\ x = \frac{(b^{2} ac) \pm \sqrt{{(b^{2} ac)}^{2} + 4 {ab}^{2} c}}{2 ab} \\ x = \frac{(b^{2} ac) \pm \sqrt{b^{4} 2 {ab}^{2} c + a^{2} c^{2} + 4 {ab}^{2} c}}{2 ab} \\ x = \frac{(b^{2} ac) \pm \sqrt{{(b^{2} + ac)}^{2}}}{2 ab} \\ x = \frac{(b^{2} ac) + (b^{2} + ac)}{2 ab}, x = \frac{(b^{2} ac) (b^{2} + ac)}{2 ab} \\ x = \frac{2 ac}{2 ab}, x = \frac{2 b^{2}}{2 ab} \\ x = \frac{c}{b}, x = \frac{b}{a} \end{array}$

Q.4 Product of two consecutive positive integers is 240. Find the integers.

Marks:1
Ans

$\begin{array}{l} Let x and (x + 1) are the consecutive integers. \\ Then, x (x + 1) = 240 \\ x^{2} + x 240 = 0 \\ x^{2} + 16 x 15 x 240 = 0 \\ x (x + 16) 15 (x + 16) = 0 \\ (x + 16) (x 15) = 0 \\ x = 16 and x = 15 \end{array}$

CBSE Important Questions Class 10 Maths Chapter 4 – Quadratic Equations

CBSE Important Questions Class 10 Maths Chapter 4 – Quadratic Equations

Related Study Material on Extramarks

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