How do you solve #x^2-75=0#?

1 Answer
Aug 17, 2016

Answer:

#x=+-5sqrt(3)#

Explanation:

Treat #x^2-75# as the difference of squares

We know that #a^2-b^2=(a-b)(a+b)#

So replacing #a# with #x#
and #b# with #5sqrt(3)# (so that #b^2=75#)
we have
#color(white)("XXX")x^2-75=0#
is equivalent to
#color(white)("XXX")(x-5sqrt(3))(x+5sqrt(3))=0#

which implies
either #(x-5sqrt(3))=0 color(white)("XX")orcolor(white)("XX") (x+5sqrt(3))=0#
#color(white)("XX")rarr x=5sqrt(3)color(white)("XXXXXXXX")rarrx=-5sqrt(3)#