How do you solve # ( x + 5)² = 49#?

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Answer 1 · 2016-06-01T12:20:15

#x=-12# or #x=2#

#(x+5)^2=49#

or #(x+5)^2-49=0#

or #(x+5)^2-7^2=0#

Now using #a^2-b^2=(a+b_(a-b)#

The above is equivalent to #((x+5)+7)((x+5)-7)=0#

or #(x+12)(x-2)=0#

Hence either #x+12=0# or #x-2=0#

i.e. #x=-12# or #x=2#