How do you solve #x^2-175=0# using the quadratic formula?

1 Answer
Jul 25, 2016

#x=+-5sqrt(7)#

Explanation:

The quadratic formula for an equation in the form:
#color(white)("XXX")color(red)(a)x^2color(blue)(+b)x+color(green)(c)=0#
gives us the solutions:
#color(white)("XXX")x=(-color(blue)(b)+-sqrt(color(blue)(b)-4(color(red)(a)color(green)(c))))/(2color(red)(a))#

#x^2-175=0# is equivalent to
#color(white)("XXX")color(red)(1)x^2+color(blue)(0)x+color(green)(""(-175))=0#

So the solutions are
#color(white)("XXX")x=(-color(blue)(0)+-sqrt(color(blue)(0)-4 * (color(red)(1) * color(green)(""(-175)))))/(2 * color(red)(1))#

#color(white)("XXXX")=+-sqrt(700)/2#

#color(white)("XXXX")=+-(10sqrt(7))/2#

#color(white)("XXXX")=+-5sqrt(7)#