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Math Help - finding the variables

  1. #1
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    finding the variables

    what would be the varibles for:

    -2x^2+6x-11=0
    I know i use a quadratic formula but i get a negative square root and dont remember what to do next since its an imaginary number.

    and

    5 times 2^x+3=35
    For this one i dont really get what to do first
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  2. #2
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    Quote Originally Posted by takkun0486
    what would be the varibles for:

    -2x^2+6x-11=0
    I know i use a quadratic formula but i get a negative square root and dont remember what to do next since its an imaginary number.
    Thus,
    x=\frac{-6\pm\sqrt{36-88}}{-4}
    Thus,
    x=\frac{-6\pm\sqrt{-52}}{-4}
    Thus,
    x=\frac{-6\pm 2i\sqrt{13}}{-4}
    Thus,
    x=\frac{3}{2}\pm \frac{\sqrt{13}}{2} i
    ------
    Ya got,
    5(2^x+3)=35
    Divide by 5,
    2^x+3=7
    Thus,
    2^x=4
    Thus,
    2^x=2^2
    Same bases thus,
    x=2
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    Thus,
    x=\frac{-6\pm\sqrt{36-88}}{-4}
    Thus,
    x=\frac{-6\pm\sqrt{-52}}{-4}
    Thus,
    x=\frac{-6\pm 2i\sqrt{13}}{-4}
    Thus,
    x=\frac{3}{2}\pm \frac{\sqrt{13}}{2} i
    ------
    Ya got,
    5(2^x+3)=35
    Divide by 5,
    2^x+3=7
    Thus,
    2^x=4
    Thus,
    2^x=2^2
    Same bases thus,
    x=2
    for the 2nd one i had the "x+3" as an exponent of 2
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  4. #4
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    Quote Originally Posted by takkun0486
    for the 2nd one i had the "x+3" as an exponent of 2
    Next time use parantheses,
    Anyways the problem reduces to,
    2^{x+3}=2^2
    Thus,
    x+3=2
    Thus,
    x=-1
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  5. #5
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    so it was
    5(2^(x+3))=35

    sry that i confused you on the other parts
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  6. #6
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    any ideas?
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  7. #7
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    Quote Originally Posted by takkun0486
    any ideas?
    Oh!
    Sorry I forgot you cannot subtract, then you have,
    2^{x+3}=7
    Take logarithms,
    \log 2^{x+3}=\log 7
    Bring down exponent bases on rule,
    (x+3)\log 2=\log 7
    Thus,
    x+3=\frac{\log 7}{\log 2}
    Thus,
    x=\frac{\log 7}{\log 2}-3\approx -.19
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