**Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations**

**Nature of Roots of a Quadratic Equation**

Let the given equation be ax^{2} + bx + c = 0, where a = 0.

Then, the discriminant is given by

D = (b^{2} – 4ac) . And, the roots of the given equation are

**Case 1:** If D > 0, then the roots are ** real and distinct**.

These roots are given by,

**Case 2:** If D = 0, then In this case, the roots are ** real and equal**.

Each root = -b/2a

**Case 3:** If D < 0, then the roots are ** imaginary**. So, the given equation has

**.**

*no real roots***Question: **Show that the equation 2×2 – 6x + 3 = 0 has real roots.

*Solution: *

The given equation is 2x^{2} – 6x + 3 = 0.

This is of the form ax^{2} + bx + c = 0, where a = 2, b = -6 and c = 3.

Therefore, D = (b^{2} – 4ac) = [(-6)^{2} – 4 x 2 x 3] = (36 – 24) = 12 > 0.

So, the given equation has real unequal roots.

**Question:** Find the value of k for which the roots of the quadratic equation kx(x – 2) + 6 = 0 are equal.

**Solution:**

The given equation is kx^{2} – 2kx + 6 = 0.

This is of the form ax^{2} + bx + c = 0, where a = k, b = -2k and c = 6.

D = (b^{2} – 4ac) = (4k^{2} – 4 x k x 6) = (4k^{2} – 24k).

For equal roots, we must have

D = 0

⇒ 4k^{2} – 24k=0

⇒ 4k(k – 6) = 0

Either k = 0 or k = 6.

Now, k ≠ 0, we get 6 = 0, which is not possible.

Therefore, k = 6.

**Q.1. **Show that the roots of the equation x^{2} + px – q^{2} = 0 are real for all real values of p and q.

**Q.2.** Find the nature of the roots of the following quadratic equations:

(i) 2x^{2} – 8x + 5 = 0 (iv) 5x(x – 2) + 6 = 0 (v) 12x^{2 }– 4√15 x + 5 = 0

Ans. (i) Real and unequal (iv) Not real (v) Real and equal

**Q.3.** For what value of k are the roots of the quadratic equation kx(x – 2√5) + 10 = 0 real and equal?

Ans. k = 0 or k = 2

**Q.4.** If the quadratic equation (1 + m^{2})x^{2} + 2mcx + c^{2} – a^{2} = 0 has equal roots, prove that c^{2} = a^{2} (1 + m^{2}) .

**Q.5.** (i) Find the values of k for which the quadratic equation (3k + 1)x^{2} + 2(k + 1)x + 1 = 0 has real and equal roots.

(ii) Find the value of k for which the equation x^{2} + k(2x + k – 1) + 2 = 0 has real and equal roots.

Ans. (i) k = 0 or k = 1 (ii) k = 2

**Q.6. **If –5 is a root of the quadratic equation 2x^{2} + px – 15 = 0 and the quadratic equation p(x^{2} + x) + k = 0 has equal roots, fi nd the value of k.

Ans. 49/28

**Q.7.** For what values of k are the roots of the quadratic equation 3x^{2} + 2kx + 27 = 0 real and equal?

Ans. k = 9 or k = -9

**Q.8.** If the roots of the equations ax^{2} + 2bx + c = 0 and bx^{2} – 2√ac x + b = 0 are simultaneously real then prove that b^{2} = ac.

**Q.9. **If –4 is a root of the equation x^{2} + 2x + 4p = 0, fi nd the value of k for which the quadratic equation x^{2} + px(1 + 3k) + 7(3 + 2k) = 0 has equal roots.

Ans. k=2 or k=-10/9

**Q.10. **Find the value of α for which the equation (α – 12)x^{2} + 2(α – 12)x + 2 = 0 has equal roots.

Ans. α=14

**Q.11. **Find the values of k for which the given quadratic equation has real and distinct roots:

(i) kx^{2} + 6x + 1 = 0

(ii) x^{2} – kx + 9 = 0

(iii) 9x^{2} + 3kx + 4 = 0

Ans. (i) k < 9 (ii) k > 6 or k < –6 (iii) k > 4 or k < –4

**Q.12.** For what values of p are the roots of the equation 4×2 + px + 3 = 0 real and equal?

Ans. p = 4√3 or p = -4√3

### Frequently Asked Questions – FAQs

## How to find roots of a quadratic equation?

There are various method to find out the roots of quadratic equation. You can use quadratic formula to find the roots of a quadratic equation.

## What is the nature of root when D=0?

The quadratic equation has only one real root (or two equal roots -b/2a and -b/2a) when D or b^{2} – 4ac = 0.

## What is the nature of roots when D<0?

The quadratic equation has two complex roots when b^{2} – 4ac < 0. The complex roots always occur in pairs.