What kind of solutions does #2x^2 + 5x + 5 = 0# have?

1 Answer
Jul 10, 2015

Answer:

#2x^2+5x+5=0# has no real roots. It has two distinct complex roots which are complex conjugates of one another.

Explanation:

#f(x) = 2x^2+5x+5# is of the form #ax^2+bx+c# with #a=2#, #b=5# and #c=5#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 5^2 - (4xx2xx5) = 25 - 40 = -15#

Since the discriminant is negative, #f(x) = 0# has no real roots. It only has complex ones.

The quadratic formula still works, giving the roots as:

#x = (-b+-sqrt(Delta))/(2a) = (-5+-sqrt(-15))/(2*2)#

#=(-5+-i sqrt(15))/4#

In general the various cases for the different values of the discriminant are as follows:

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

#Delta = 0# The quadratic equation has one repeated real root. It is a perfect square trinomial.

#Delta < 0# The quadratic equation has no real roots. It has a conjugate pair of distinct complex roots.