# Discriminant

How to find the nature of the roots of a quadratic equation by using its discriminant: formula, 4 examples, and their solutions.

## Formula

For a quadratic equation

ax^{2} + bx + c = 0

(a ≠ 0),

x = [-b ± √b^{2} - 4ac] / 2a.

Quadratic Formula

The discriminant, D, is the number

inside the radical sign:

b^{2} - 4ac.

## Nature of the Roots

D = b^{2} - 4ac determines

the nature of the roots.

So, without solving the quadratic equation,

you can check the nature of the roots

by finding D.

If D is plus and is a perfect square,

then it has two rational roots.

If D is plus and is not a perfect square,

then it has two irrational roots.

If D = 0,

then it has one real root.

If D is minus (< 0),

then it has no real roots.

(= two complex roots)

## Examplex^{2} + 7x + 10 = 0

The given quadratic equation is

1x^{2} + 7x + 10 = 0.

a = 1

b = +7

c = +10

Then D = 7^{2} - 4⋅1⋅10.

7^{2} = 49

-4⋅1⋅10 = -40

49 - 40 = 9 = 3^{2}

D = 3^{2}

D is plus.

And D is a perfect square.

Then the quadratic equation has

two rational roots.

So

two rational roots

is the answer.

## Examplex^{2} - 4x - 1 = 0

The given quadratic equation is

1x^{2} - 4x - 1 = 0.

a = 1

b = -4

c = -1

Then D = (-4)^{2} - 4⋅1⋅(-1).

(-4)^{2} = 16

-4⋅1⋅(-1) = +4

16 + 4 = 20

D = 20

D is plus.

And D is not a perfect square.

Then the quadratic equation has

two irrational roots.

So

two irrational roots

is the answer.

## Examplex^{2} - 6x + 9 = 0

The given quadratic equation is

1x^{2} - 6x + 9 = 0.

a = 1

b = -6

c = +9

Then D = (-6)^{2} - 4⋅1⋅9.

(-6)^{2} = 36

-4⋅1⋅9 = -36

36 - 36 = 0

D = 0

Then the quadratic equation has

one real root.

So

one real root

is the answer.

## Examplex^{2} + 2x + 5 = 0

The given quadratic equation is

1x^{2} + 2x + 5 = 0.

a = 1

b = +2

c = +5

Then D = 2^{2} - 4⋅1⋅5.

2^{2} = 4

-4⋅1⋅5 = -20

4 - 20 = -16

D = -16

D is minus.

Then the quadratic equation has

no real roots.

So

no real roots

is the answer.