How do you solve #(b+2)^2 = 4#?

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Answer 1 · 2018-03-28T21:14:51

#b=0#

Take the square root of both sides to eliminate the square:

#sqrt((b+2)^2)=sqrt4#

so you get:

#b+2=2#

Subtract 2 from both sides, which gives you your answer:

#b=0#