How do you solve #x^2+8=0# using the quadratic formula?

1 Answer
Jan 16, 2017

Answer:

#x=2sqrt(2)color(white)("x")icolor(white)("XXX")orcolor(white)("XXX")x=-2sqrt(2)color(white)("x")i#

(see below for use of the quadratic formula)

Explanation:

Standard quadratic form for an equation is:
#color(white)("XXX")color(red)ax^2+color(blue)bx+color(magenta)c=0#
and in this form the solutions are given by
#color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(magenta)c))/(2color(red)a)#

Expressing the given equation into standard form:
#x^2+8=0#
#color(white)("XXX")rarr color(red)1x^2+color(blue)0 * x +color(magenta)8 = 0#
with solutions:
#color(white)("XXX")x=(-color(blue)0+-sqrt(color(blue)0^2-4 * color(red)1 * color(magenta)8))/(2 * color(red)1)#

#color(white)("XXX")=+-sqrt(-32)/2=+-(4sqrt(2)color(white)("x")i)/2=+-2sqrt(2)color(white)("x")i#