What is #x# in #(x-2)^2=64# ?

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Answer 1 · 2018-05-24T23:04:26

#x=-6 or x=10#

#(x - 2)^2 = 64#

The exponent #2# means that #x-2# will multiply itself twice.

#(x - 2)(x - 2) = 64#

Use the distributive property on the left side

#(x)(x) + (x)(-2) + (-2)(x) + (-2)(-2)=64#

#x^2 - 2x - 2x +4 = 64#

#x^2 - 4x + 4 = 64#

Now we can subtract #64# from both sides

#x^2 - 4x + 4 - 64 = 64 - 64#

#x^2 - 4x - 60 = 0#

Then factorize the left side

#(x + 6)(x-10)=0#

Now we can set the factors equal to #0#

#x + 6 = 0 or x - 10 = 0#

#x = 0 - 6 or x = 0 + 10#

#x = -6 or x = 10# #larr# This is the final answer!