Nature of roots — lesson. Mathematics State Board, 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

$ax^2 + bx + c = 0$ . It is denoted by the letter

$\Delta$ or

$D$ .

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

Case I:

$\Delta = b^2 - 4ac > 0$

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

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

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

Case II:

$\Delta = b^2 - 4ac = 0$

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

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

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

Case III:

$\Delta = b^2 - 4ac < 0$

Here,

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

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

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

Relation between roots and coefficients of a quadratic equation

$\alpha$ and

$\beta$ are the roots of the quadratic equation

$ax^2 + bx + c = 0$ , then:

(i) Sum of the roots

$=$

$α + β = \frac{- b}{a}$

(ii) Product of the roots

$=$

$α β = \frac{c}{a}$

Quadratic equation

$=$

$x^2 - (\text{sum of the roots})x + \text{product of the roots}$

Important!

To know more about the relationship between roots and coefficients of a equation click here.

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1. Nature of roots

Theory: