How do you use the discriminant to find the number of real solutions of the following quadratic equation: #2x^2 + 4x + 2 = 0#?

1 Answer
Jun 27, 2015

Answer:

#b^2-4ac = 4^2 - 4(2)(2) = 0# so this equation has exactly one real solution.

Explanation:

For a quadratic of the form #ax^2+bx+c = 0#
the solution roots are given by the quadratic formula: #x=(-b+-sqrt(b^2-4ac))/(2a)#

The discriminant is the portion #(b^2-4ac)#
if it is negative then the roots include a term which is the square root of a negative number; since there are no Real numbers whose square root is negative, when the discriminant is #<0# there are no Real roots.

If the discriminant is #=0# then #+-sqrt(0)# is a single value (#+0 = -0#)
and the quadratic has exactly one Real solution.

If the discriminant is #>0# then #+-sqrt("discriminant")# represents two different values and the quadratic has two Real solutions.