How do you solve #r^2+25=0# using the quadratic formula?

1 Answer
Dec 3, 2016

Answer:

#r=-5i or r=+5i#

Explanation:

The quadratic formula tells us that a quadratic in standard form:
#color(white)("XXX")color(red)ax^2+color(blue)bx+color(green)c=0#
has solutions given by:
#color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(green)c))/(2color(red)a)#

Converting the given: #r^2+25=0# into standard form:
#color(white)("XXX")color(red)1r^2+color(blue)0r+color(green)(25)=0#
we have the solutions:
#color(white)("XXX")r=(color(blue)0+-sqrt(color(blue)0^2-4*color(red)1*color(green)(25)))/(2*color(red)1)#

#color(white)("XXX")=+-(2sqrt(-25))/2 = +-sqrt(-25) = +-5i#