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36

\[A=x ^{2}\]
with x being the length of the initial sides
\[A+64=(x+4)\]

the area of previous square was 36 sq. inches

\[(x+4)^{2}\]
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

hmm

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