How do you solve #(3x+1)^2+(4x-3)^2=(5x-1)^2#?

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Answer 1 · 2017-04-26T21:08:23

#x=9/8#

Given:

#(3x+1)^2+(4x-3)^2=(5x-1)^2#

Multiply out both sides to get:

#9x^2+6x+1+16x^2-24x+9=25x^2-10x+1#

Simplifying the left hand side, this becomes:

#color(red)(cancel(color(black)(25x^2)))-18x+10 = color(red)(cancel(color(black)(25x^2)))-10x+1#

Add #-25x^2+18x-1# to both sides to get:

#9=8x#

Divide both sides by #8# to get:

#x = 9/8#