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.