Given the equation #(ax+b)(ax-b)=0#, how do you solve for x?

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Answer 1 · 2016-12-21T16:17:10

#x=+-b/a#

multiply out and rearrange for #""x#

#(ax+b)(ax-b)#

this is difference of squares

#a^2x^2-b^2=0#

#a^2x^2=b^2#

#x^2=b^2/a^2#

#x=+-sqrt(b^2/a^2#

#x=+-b/a#