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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 ax2+bx+c = 0 where a, b, c R & a 0 , then

x = b±D2a    where D = b24ac

So, a quadratic equation ax2+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 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 R & i = 1).

(b) Consider the quadratic equation ax2+bx+c = 0 where a, b, c Q & a 0, then :

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

(ii) if α = p + q is one root in this case, (where p is rational & q is a surd) then other root will be p – q.

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

(i) x2+x+1 = 0

(ii) 3x2+2x1 = 0

Solution

(i) We have, x2+x+1 = 0

Here,  a = 1,  b = 1,  c = 1

Therefore, Discriminant D = b24ac = (1)24(1)(1) = -3 < 0.

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

(ii) We have, 3x2+2x1 = 0

Here,  a = 3,  b = 2,  c = -1

Therefore, Discriminant D = b24ac = (2)24(3)(1) = 16 > 0.

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

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