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Thread: Rational Exponent Equation.

  1. #1
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    Rational Exponent Equation.

    I am having some trouble with this one. I think exponentiation would work but i can't seem to get it right. the answer is x=36. could someone show an intermediate step so i can see which form, radical or fractional exponent to use. thanks.

    $\displaystyle

    {x}^{\frac {1} {2}}-6{x}^{\frac {-1}{2}}-5=0

    $

    $\displaystyle

    \sqrt {x}-6\,{\frac {1} {\sqrt {x}}}}-5=0}

    $
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  2. #2
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    e^(i*pi)'s Avatar
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    Multiply through by $\displaystyle x^{1/2}$

    $\displaystyle x - 6 - 5\sqrt{x} = 0$ or, in standard form: $\displaystyle (\sqrt{x})^2 - 5\sqrt{x} - 6 =0$

    This is a quadratic equation (and factors)


    Edit: Note that the domain demands that $\displaystyle x > 0$ so discard any negative solution
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    I tried to do a normal quadratic factor and get (x=4,x=9) which is not right. quadratic formula does not seem to give 36 also. I also tried to exponentiation but it just shifts the radical on x. how to factor this?
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    e^(i*pi)'s Avatar
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    $\displaystyle (\sqrt{x}+1)(\sqrt{x}-6) = 0$ - using the quadratic formula would give you $\displaystyle \sqrt{x}$ so you'd need to square the value obtained to get x. Hence if you got 6 from the quadratic formula you'd have done it right

    Perhaps it's easier if you let $\displaystyle u = \sqrt{x}$ which would give $\displaystyle u^2-5u-6=0$
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  5. #5
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    From $\displaystyle (\sqrt{x})^2 - 5\sqrt{x} - 6 =0$

    make $\displaystyle u= \sqrt{x}$

    $\displaystyle u^2 - 5u - 6 =0$

    $\displaystyle (u-6)(u+1) = 0$

    $\displaystyle (\sqrt{x}-6)(\sqrt{x}+1) = 0$

    Finish him off!
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  6. #6
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    ah I got it thanks. I got $\displaystyle (\sqrt{x}-2)(\sqrt{x}-3) = 0$ which looked right for a second. Is there a rule where you can't get 2 negative factors if the quadratic has negative terms, or something like that?
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    e^(i*pi)'s Avatar
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    That's not right. The last term will be +6 instead of -6.

    Me and Pickslides have given the correct factorisation in posts 4 and 5 respectively although I use trial and error to get it right when factoring
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  8. #8
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    yah. you are correct thanks for you help.
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  9. #9
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    You will have two values for $\displaystyle \sqrt{x}$.
    You need to get rid of the negative one because for your purposes we say that the square root of a negative number is undefined. So there is one correct value for $\displaystyle \sqrt{x}$ here.

    I think that's what you were asking about, but I'm not sure.
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