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Soni-kay

  • 3 years ago

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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  1. amruta1997
    • 3 years ago
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    36

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

  3. amruta1997
    • 3 years ago
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    the area of previous square was 36 sq. inches

  4. AlexUL
    • 3 years ago
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    \[(x+4)^{2}\] sorry

  5. tcarroll010
    • 3 years ago
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    (x + 4)^2 = x^2 + 64. This will be a quadratic in "x".

  6. Soni-kay
    • 3 years ago
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    how did you get 36 so i can do the rest by myself?

  7. amruta1997
    • 3 years ago
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    it can be done by trial and error also

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

  9. Soni-kay
    • 3 years ago
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    ok one thing why do you square it

  10. amruta1997
    • 3 years ago
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    area of square is its side squared

  11. tcarroll010
    • 3 years ago
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    You have to get the area. Both of the first square and the second.

  12. amruta1997
    • 3 years ago
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    and x is the side

  13. Soni-kay
    • 3 years ago
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    thanks guys

  14. tcarroll010
    • 3 years ago
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    The second area is (x + 4)^2 and the first area is x^2

  15. amruta1997
    • 3 years ago
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    hmm

  16. amruta1997
    • 3 years ago
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    its our pleasure

  17. AlexUL
    • 3 years ago
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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?

  18. Soni-kay
    • 3 years ago
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    lol i got it

  19. Soni-kay
    • 3 years ago
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    ok so i did x^2 + 8x + 16 = x^2 + 64. and i got 6 what did i do wrong

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

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