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Sonikay
Group Title
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?
 2 years ago
 2 years ago
Sonikay Group Title
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?
 2 years ago
 2 years ago

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AlexUL Group TitleBest ResponseYou've already chosen the best response.2
\[A=x ^{2}\] with x being the length of the initial sides \[A+64=(x+4)\]
 2 years ago

amruta1997 Group TitleBest ResponseYou've already chosen the best response.0
the area of previous square was 36 sq. inches
 2 years ago

AlexUL Group TitleBest ResponseYou've already chosen the best response.2
\[(x+4)^{2}\] sorry
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
(x + 4)^2 = x^2 + 64. This will be a quadratic in "x".
 2 years ago

Sonikay Group TitleBest ResponseYou've already chosen the best response.0
how did you get 36 so i can do the rest by myself?
 2 years ago

amruta1997 Group TitleBest ResponseYou've already chosen the best response.0
it can be done by trial and error also
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
(x + 4)^2 = x^2 + 64 will become x^2 + 8x + 16 = x^2 + 64. Just combine terms and simplify.
 2 years ago

Sonikay Group TitleBest ResponseYou've already chosen the best response.0
ok one thing why do you square it
 2 years ago

amruta1997 Group TitleBest ResponseYou've already chosen the best response.0
area of square is its side squared
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
You have to get the area. Both of the first square and the second.
 2 years ago

amruta1997 Group TitleBest ResponseYou've already chosen the best response.0
and x is the side
 2 years ago

Sonikay Group TitleBest ResponseYou've already chosen the best response.0
thanks guys
 2 years ago

tcarroll010 Group TitleBest ResponseYou've already chosen the best response.0
The second area is (x + 4)^2 and the first area is x^2
 2 years ago

amruta1997 Group TitleBest ResponseYou've already chosen the best response.0
its our pleasure
 2 years ago

AlexUL Group TitleBest ResponseYou've already chosen the best response.2
\[A _{1}=x _{1}^{2}\] \[A _{2}=x _{2}^{2}\] \[A _{2}=A _{1}+64\] \[x _{2}=x _{1}+4\] Does that help?
 2 years ago

Sonikay Group TitleBest ResponseYou've already chosen the best response.0
lol i got it
 2 years ago

Sonikay Group TitleBest ResponseYou've already chosen the best response.0
ok so i did x^2 + 8x + 16 = x^2 + 64. and i got 6 what did i do wrong
 2 years ago

AlexUL Group TitleBest ResponseYou've already chosen the best response.2
you got \[x _{1}=6\] so what's the area? Never forget what was the initial question
 2 years ago
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