Nature of roots — lesson. Mathematics CBSE, Class 10.

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We have learnt that the roots of the quadratic equation

$ax^2 + bx + c = 0$ can be found by the quadratic formula:

$x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}$

Important!

$b^2 - 4ac$ is called the discriminant of the quadratic equation

$ax^2 + bx + c = 0$ .

Let us discuss the nature of the roots of the quadratic equation depending on the discriminant.

Case I:

$b^2 - 4ac > 0$

Here,

$b^2 - 4ac > 0$ . That means the value of the discriminant is positive.

Then, the possible roots are

$\frac{- b + \sqrt{b^{2} - 4 ac}}{2 a}$ and

$\frac{- b - \sqrt{b^{2} - 4 ac}}{2 a}$ .

$b^2 - 4ac > 0$ , then the roots are real and distinct.

Case II:

$b^2 - 4ac = 0$

Here,

$b^2 - 4ac = 0$ . That means the value of the discriminant is zero.

$x = \frac{- b + \sqrt{0}}{2 a}$ and

$x = \frac{- b - \sqrt{0}}{2 a}$

$x = \frac{- b}{2 a}$ and

$x = \frac{- b}{2 a}$

The possible roots are

$\frac{- b}{2 a}$ and

$\frac{- b}{2 a}$ .

$b^2 - 4ac = 0$ , then the roots are real and equal.

Case III:

$b^2 - 4ac < 0$

Here,

$b^2 - 4ac < 0$ . That means the value of the discriminant is negative.

We won't get any real roots in this case.

$b^2 - 4ac < 0$ , then there are no real roots.

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1. Nature of roots