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

1 Answer
Jul 4, 2015

Answer:

#2x^2+2x+2=0# has discriminant #Delta = -12# which is negative. So there are no real solutions, only two distinct complex ones.

Explanation:

#2x^2+2x+2# is of the form #ax^2+bx+c# with #a=2#, #b=2# and #c=2#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 2^2-(4xx2xx2) = 4 - 16 = -12#

Since #Delta < 0# there are no real solutions of #2x^2+2x+2=0#. It has two distinct complex solutions.

The possibilities are:

#Delta > 0# The quadratic has two distinct solutions. If #Delta# is a perfect square (and the coefficients of the quadratic are rational) then the roots are rational too.

#Delta = 0# The quadratic has one repeated real root.

#Delta < 0# The quadratic has no real roots. It has a pair of complex roots which are conjugates of one another.