# Nature of Roots of Quadratic Equation

Here you will learn how to find the nature of roots of quadratic equation using discriminant with examples.

Let’s begin –

## Nature of Roots of Quadratic Equation

(a) Consider the quadratic equation $$ax^2 + bx + c$$ = 0 where a, b, c $$\in$$ R & a $$\ne$$ 0 , then

x = $$-b \pm \sqrt{D}\over 2a$$    where D = $$b^2 – 4ac$$

So, a quadratic equation $$ax^2 + bx + c$$ = 0

(i) has no real roots if D < 0.

(ii) has two equal real roots if D = 0.

(iii) has two distinct real roots if D > 0.

(iv) has real roots if D $$\ge$$ 0.

(v) If p + iq is one of the root of a quadratic equation, then the other root must be the conjugate p – iq & vice versa. (p, q $$\in$$ R & i = $$\sqrt{-1}$$).

(b) Consider the quadratic equation $$ax^2 + bx + c$$ = 0 where a, b, c $$\in$$ Q & a $$\in$$ 0, then :

(i) If D is a perfect square, then roots are rational.

(ii) if $$\alpha$$ = p + $$\sqrt{q}$$ is one root in this case, (where p is rational & $$\sqrt{q}$$ is a surd) then other root will be p – $$\sqrt{q}$$.

Example : Examine, whether the following equations have real roots :

(i) $$x^2 + x + 1$$ = 0

(ii) $$3x^2 + 2x – 1$$ = 0

Solution

(i) We have, $$x^2 + x + 1$$ = 0

Here,  a = 1,  b = 1,  c = 1

Therefore, Discriminant D = $$b^2 – 4ac$$ = $$(1)^2 – 4(1)(1)$$ = -3 < 0.

Since, D < 0, the equation has no real roots.

(ii) We have, $$3x^2 + 2x – 1$$ = 0

Here,  a = 3,  b = 2,  c = -1

Therefore, Discriminant D = $$b^2 – 4ac$$ = $$(2)^2 – 4(3)(-1)$$ = 16 > 0.

Since, D > 0, the equation has real roots.