# DISCUSS THE NATURE OF ROOTS OF A QUADRATIC EQUATION

Discuss the Nature of Roots of a Quadratic Equation :

In this section, you will learn, how to examine the nature of the roots of a quadratic equation.

## Discuss the Nature of Roots of a Quadratic Equation - Practice Problems with Solutions

Problem 1 :

If the equations x2 −ax+b = 0 and x2 −ex+f = 0 have one root in common and if the second equation has equal roots, then prove that ae = 2(b + f).

Solution :

Let "α" be the common root for both quadratic equations

Let "β" be the other root of the quadratic equations

Since the roots of the second equation will be same, "α" and "α" are the roots of the second equation.

 x2 − ax + b = 0Sum of roots  =  aα + β  =  a --(1)Product of roots αβ  =  bβ  =  b/α --(2) x2 − ex + f = 0Sum of roots  =  eα + α  =  e2α = e  ==>  α = e/2--(3)Product of roots α(α)  =  fα2 = f --(4)

Problem 2 :

Discuss the nature of roots of

−x2 + 3x + 1 = 0

Solution :

To find the nature of roots, we have to use the formula for discriminant.

Discriminant  =  b2 - 4ac

a  =  -1, b = -3 and c  =  1

=  (-3)2 - 4(-1) (1)

=  9 + 4

=  13 > 0

So, the roots are real and distinct.

Problem 3 :

Discuss the nature of roots of

4x2 − x − 2 = 0

Solution :

To find the nature of roots, we have to use the formula for discriminant.

Discriminant  =  b2 - 4ac

a  =  4, b = -1 and c  =  -2

=  (-1)2 - 4(4) (-2)

=  1 + 64

=  65 > 0

So, the roots are real and distinct.

Problem 4 :

Discuss the nature of roots of

9x2 + 5x = 0.

Solution :

To find the nature of roots, we have to use the formula for discriminant.

Discriminant  =  b2 - 4ac

a  =  9, b = 5 and c  =  0

=  (5)2 - 4(9) (0)

=  25 > 0

So, the roots are real and distinct. After having gone through the stuff given above, we hope that the students would have understood, how to examine the nature of roots of a quadratic equation

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