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

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

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