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anonymous

  • 4 years ago

Solve x2 + 8x – 48 = 0 by completing the square. Show your work for full credit.

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  1. radar
    • 4 years ago
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    For completing the square the coefficient of the "squared" term should be 1 and it is.

  2. radar
    • 4 years ago
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    The term to be added to make it a perfect square is one half of the coefficent of the x term squared.

  3. radar
    • 4 years ago
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    What you add you must also subtract to retain the equations original value.

  4. anonymous
    • 4 years ago
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    \[(x^2+8x+16)-16-48\] <---- just added and substracted 4, which is \[(8/2)^2\] Now try to factor out the term in brackets what do you get?

  5. radar
    • 4 years ago
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    Lets see how saljudieh07 did this

  6. radar
    • 4 years ago
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    Looks good.

  7. anonymous
    • 4 years ago
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    sorry \[4^2\] that is what I added and substracted

  8. anonymous
    • 4 years ago
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    now, kaymarie12479 can you factor the expression in the brackets?

  9. radar
    • 4 years ago
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    Actually you want to the square root of the value in the brackets. The origiinal equation is factorable, but you want to "complete the square. \[(x + 4)^{2}=64\]

  10. radar
    • 4 years ago
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    \[x+4=\pm8\]

  11. radar
    • 4 years ago
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    Note the original equation is factorable (x-4)(x+12)=0 however, they want you to use completing the square method.

  12. radar
    • 4 years ago
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    Note. That factoring gives you x=4 and x=-12 as answer and x + 4 = +/- 8 gives same results x=4, and x=-12

  13. radar
    • 4 years ago
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    @saljudieh07 you left off the "=0" and disregarded your constants.

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