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}

If \(b^2 - 4ac > 0\), 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}

If \(b^2 - 4ac = 0\), 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.

If \(b^2 - 4ac < 0\), then there are no real roots.

Did you find an error?

1. Nature of roots