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

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Answer 1 · 2016-11-06T03:50:43

#(2x - 3)^2 = x^2 - 3#

#(2x - 3)(2x - 3) = x^2 - 3#

#4x^2 - 6x - 6x + 9 = x^2 - 3#

#4x^2 - 12x + 9 = x^2 - 3#

#3x^2 - 12x + 12 = 0#

#3(x^2 - 4x + 4) = 0#

#x^2 - 4x+ 4 = 0#

#(x - 2)(x- 2) = 0#

#x= 2#

Check:

#(2(2) - 3)^2 =^? 2^2 - 3#

#1^2 =1" " color(green)√#

Hopefully this helps!