How do you solve #x^2=16# using the quadratic formula?

1 Answer
Jan 5, 2017

Answer:

(see below for solution using quadratic formula)
#color(white)("XXX")x=+-4#

Explanation:

Rewriting #x^2=16# into explicit standard quadratic form:
#color(white)("XXX")color(red)1x^2+color(blue)0x+color(green)(""(-16))=0#

The quadratic formula tells us that a quadratic equation with the form:
#color(white)("XXX")color(red)ax^2+color(blue)bx+color(green)c=0#
has solutions:
#color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b^2-4 * color(red)a * color(green)c))/(2 * color(red)a)#

Our (standardized) quadratic has
#color(white)("XXX")color(red)a=color(red)1#
#color(white)("XXX")color(blue)b=color(blue)0#
#color(white)("XXX")color(green)c=color(green)(""(-16))#

So its solutions are
#color(white)("XXX")x=(color(blue)0+-sqrt(color(blue)0^2-4 * color(red)1 * color(green)(""(-16))))/(2 * color(red)1#

#color(white)("XXXX")=+-sqrt(64)/2#

#color(white)("XXXX")=+-8/2#

#color(white)("XXXX")=+-4#