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When 4 inches are added to each side of square, its area increases by 64 square inches. What is the area of the initial square?

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\[A=x ^{2}\] with x being the length of the initial sides \[A+64=(x+4)\]
the area of previous square was 36 sq. inches

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Other answers:

\[(x+4)^{2}\] sorry
(x + 4)^2 = x^2 + 64. This will be a quadratic in "x".
how did you get 36 so i can do the rest by myself?
it can be done by trial and error also
(x + 4)^2 = x^2 + 64 will become x^2 + 8x + 16 = x^2 + 64. Just combine terms and simplify.
ok one thing why do you square it
area of square is its side squared
You have to get the area. Both of the first square and the second.
and x is the side
thanks guys
The second area is (x + 4)^2 and the first area is x^2
its our pleasure
\[A _{1}=x _{1}^{2}\] \[A _{2}=x _{2}^{2}\] \[A _{2}=A _{1}+64\] \[x _{2}=x _{1}+4\] Does that help?
lol i got it
ok so i did x^2 + 8x + 16 = x^2 + 64. and i got 6 what did i do wrong
you got \[x _{1}=6\] so what's the area? Never forget what was the initial question

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