Discriminant

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

Formula

Formula

For a quadratic equation
ax2 + bx + c = 0
(a ≠ 0),

x = [-b ± √b2 - 4ac] / 2a.

Quadratic Formula

The discriminant, D, is the number
inside the radical sign:
b2 - 4ac.

Nature of the Roots

D = b2 - 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.

Example 1

Example

Solution

The given quadratic equation is
1x2 + 7x + 10 = 0.

a = 1
b = +7
c = +10

Then D = 72 - 4⋅1⋅10.

72 = 49

-4⋅1⋅10 = -40

49 - 40 = 9 = 32

D = 32

D is plus.
And D is a perfect square.

Then the quadratic equation has
two rational roots.

So [two rational roots] is the answer.

Example 2

Example

Solution

The given quadratic equation is
1x2 - 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.

Example 3

Example

Solution

The given quadratic equation is
1x2 - 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.

Example 4

Example

Solution

The given quadratic equation is
1x2 + 2x + 5 = 0.

a = 1
b = +2
c = +5

Then D = 22 - 4⋅1⋅5.

22 = 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.