**Nature Of The Roots Of A Quadratic Equation**

The nature of the roots depends on the value of b^{2} – 4ac. bx^{2} – 4ac is called the **discriminant** of the quadratic equation ax^{2} + bx + c = 0 and is generally, denoted by D.

∴ D = b^{2} – 4ac

**If D > 0**, i..e., b^{2} – 4ac > 0, i.e., b2 – 4ac is positive; the roots are **real and unequal**. Also,

(i) If b^{2} – 4ac is a perfect square, the roots are **rational and unequal**.

(ii) If b^{2} – 4ac is positive but not perfect square, the roots are** irrational and unequal**.

**If D = 0**, i.e., b^{2} – 4ac = 0; the roots are** real and equal**.

**If D < 0**, i.e., b2 – 4ac < 0; i.e., b^{2} – 4ac is negative; the roots are not real, i.e., the roots are** imaginary**.

**Nature Of The Roots Of A Quadratic Equation With Examples**

**Example 1:** Without solving, examine the nature of roots of the equation 2x^{2} + 2x + 3 = 0.

**Sol.** Comparing 2x^{2} + 2x + 3 = 0 with ax^{2} + bx + c = 0

we get: a = 2, b = 2 and c = 3

D = b^{2} – 4ac = (2)^{2} – 4 × 2 × 3 = 4 – 24

= – 20; which is negative.

∴ The roots of the given equation are imaginary.

**Example 2:** Without solving, examine the nature of roots of the equation 2x^{2} – 7x + 3 = 0.

**Sol.** Comparing 2x^{2} – 7x + 3 = 0 with ax^{2} + bx + c = 0;

we get: a = 2, b = – 7 and c = 3

D = b^{2} – 4ac = (–7)^{2} – 4 × 2 × 3

= 49 – 24 = 25, which is perfect square.

∴ The roots of the given equation are rational and unequal.

**Example 3:** Without solving, examine the nature of roots of the equation x^{2} – 5x – 2 = 0.

**Sol.** Comparing x^{2} – 5x – 2 = 0 with ax^{2} + bx + c = 0;

we get: a = 1, b = – 5 and c = – 2

D = b^{2} – 4ac = (–5)^{2} – 4 × 1 × – 2

= 25 + 8 = 33 ; which is positive but not a perfect square.

∴ The roots of the given equation are irrational and unequal.

**Example 4:** Without solving, examine the nature of roots of the equation 4x^{2} – 4x + 1 = 0.

**Sol.** Comparing 4x^{2} – 4x + 1 = 0 with ax^{2} + bx + c = 0;

we get: a = 4, b = – 4, and c = 1

D = b^{2} – 4ac = (–4)^{2} – 4 × 4 × 1

= 16 – 16 = 0

∴ Roots are real and equal.

**Example 5: **For what value of m, are the roots of the equation (3m + 1)x^{2} + (11 + m) x + 9 = 0 equal?

**Sol. **Comparing the given equation with ax^{2} + bx + c = 0;

we get : a = 3m + 1, b = 11 + m and c = 9

∴ Discriminant, D = b^{2} – 4ac

= (11 + m)^{2} – 4(3m + 1) × 9

= 121 + 22m + m^{2} – 108m – 36

= m^{2} – 86m + 85

= m^{2} – 85m – m + 85

= m(m – 85) – 1 (m – 85)

= (m – 85) (m – 1)

Since the roots are equal, D = 0

⇒ (m – 85) (m – 1) = 0

⇒ m – 85 = 0 or m – 1 = 0

⇒ m = 85 or m = 1

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