How do you solve $(2x-3)^2 = 8$ ?

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Answer 1 · 2016-09-01T17:31:53

$x = 2.914 or x= 0.0585$

Method 1

Find the square root of each side:

$(2x-3)^2 = 8$

$2x-3 = +-sqrt8$

Consider each root separately:

$2x-3 = sqrt8$

$x = (sqrt8+3)/2 = 2.914$

Or

$2x-3 = -sqrt8$

$x = (-sqrt8 +3)/2 = 0.0858$

Method 2

Find the square of the binomial:

$(2x-3)^2 =8$
$4x^2 -12x+9 -8 =0 larr$ make equal to 0

$4x^2 -12x +1 =0$

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$x = (-(-12)+-sqrt((-12)^2-4(4)(1)))/(2(4)$

$x =(12+-sqrt(128))/(8)$

$x = 2.914 or x = 0.0858$

How do you solve (2x-3)^2 = 8(2x-3)^2 = 8?