How do you solve ${(3 x - 2)}^{2} = {(2 x + 6)}^{2}$ $(3x-2)^2=(2x+6)^2$ ?

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Answer 1 · 2017-01-19T18:09:11

$x = 8, - \frac{4}{5}$ $x=8, -4/5$

Taking square root on both sides, $3 x - 2 = \pm (2 x + 6)$ $3x-2= +- (2x+6)$

First 3x-2= 2x+6 would give x= 8

Next 3x-2= - (2x+6) would give $x = - \frac{4}{5}$ $x= -4/5$

How do you solve (3x-2)^2=(2x+6)^2(3x−2)2=(2x+6)2(3x-2)^2=(2x+6)^2?