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.