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

1 Answer
Sep 1, 2016

Answer:

#x = 2.914 or x= 0.0585#

Explanation:

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#

Now using the quadratic formula

#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#