What is the discriminant of #x^2 + 25 = 0# and what does that mean?

1 Answer
Jul 18, 2015

Answer:

#x^2+25 = 0# has discriminant #-100 = -10^2#

Since this is negative the equation has no real roots. Since it is negative of a perfect square it has rational complex roots.

Explanation:

#x^2+25# is in the form #ax^2+bx+c#, with #a=1#, #b=0# and #c=25#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = 0^2 - (4xx1xx25) = -100 = -10^2#

Since #Delta < 0# the equation #x^2+25 = 0# has no real roots. It has a pair of distinct complex conjugate roots, namely #+-5i#

The discriminant #Delta# is the part under the square root in the quadratic formula for roots of #ax^2+bx+c = 0# ...

#x = (-b +-sqrt(b^2-4ac))/(2a) = (-b +-sqrt(Delta))/(2a)#

So if #Delta > 0# the equation has two distinct real roots.

If #Delta = 0# the equation has one repeated real root.

If #Delta < 0# the equation has no real roots, but two distinct complex roots.

In our case the formula gives:

#x = (-0 +-10i)/2 = +-5i#