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Soni-kay 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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  1. amruta1997 Group Title
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    36

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

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

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

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

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

    • 2 years ago
  7. amruta1997 Group Title
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    it can be done by trial and error also

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

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

    • 2 years ago
  10. amruta1997 Group Title
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    area of square is its side squared

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

    • 2 years ago
  12. amruta1997 Group Title
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    and x is the side

    • 2 years ago
  13. Soni-kay Group Title
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    thanks guys

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

    • 2 years ago
  15. amruta1997 Group Title
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    hmm

    • 2 years ago
  16. amruta1997 Group Title
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    its our pleasure

    • 2 years ago
  17. AlexUL Group Title
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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?

    • 2 years ago
  18. Soni-kay Group Title
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    lol i got it

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

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

    • 2 years ago
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