Question

For the quadratic equation x squared plus 3 x plus 5 equals 0, find the value of the discriminant to determine if the equation has a real or non-real solution.

Answer 1

For quadraic equation:

a x² + b x +c

∆ = b² - 4 a c

is the discriminant

The discriminant is used to determine the number of solutions in the quadratic equation.

There are three cases:

If ∆ < 0 the equation has two conjugate complex solutions

If ∆ = 0 the equation has one real solution

If ∆ > 0 the equation has two non-real (conjugate complex solutions)


In this case:

x² + 3 x + 5

The coefficients are:

a = 1 , b = 3 , c = 5

∆ = b² - 4 a c = 3² - 4 ∙ 1 ∙ 5 = 9 - 20 = - 11

∆ < 0

So the equation x² + 3 x + 5 has two non-real (conjugate complex solutions).

Answer 2

My typo.

If ∆ < 0 the equation has two non-real (conjugate complex solutions)

If ∆ = 0 the equation has one real solution

If ∆ > 0 the equation has two real solutions