Sonikay
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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amruta1997
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36

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

amruta1997
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the area of previous square was 36 sq. inches

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

tcarroll010
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(x + 4)^2 = x^2 + 64. This will be a quadratic in "x".

Sonikay
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how did you get 36 so i can do the rest by myself?

amruta1997
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it can be done by trial and error also

tcarroll010
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(x + 4)^2 = x^2 + 64 will become x^2 + 8x + 16 = x^2 + 64. Just combine terms and simplify.

Sonikay
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ok one thing why do you square it

amruta1997
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area of square is its side squared

tcarroll010
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You have to get the area. Both of the first square and the second.

amruta1997
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and x is the side

Sonikay
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thanks guys

tcarroll010
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The second area is (x + 4)^2 and the first area is x^2

amruta1997
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hmm

amruta1997
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its our pleasure

AlexUL
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\[A _{1}=x _{1}^{2}\]
\[A _{2}=x _{2}^{2}\]
\[A _{2}=A _{1}+64\]
\[x _{2}=x _{1}+4\]
Does that help?

Sonikay
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lol i got it

Sonikay
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ok so i did x^2 + 8x + 16 = x^2 + 64. and i got 6 what did i do wrong

AlexUL
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you got \[x _{1}=6\]
so what's the area?
Never forget what was the initial question