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Thread: algebra 2 trouble....

  1. #1
    Junior Member eah1010's Avatar
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    Exclamation algebra 2 trouble....

    ok there are a few that im have a lot of difficulty with.... some help would be GREAT (:


    1.) square root of (x + 6) - square root of (x-5) = 1



    2.) square root of (2x -1) = square root of (x-1) + 1



    3.) x^2/3 + x^1/3 - 2 = 0



    4.) cube root of (3x-7) = cube root of (4-x)



    5.) square root of (y) + fourth root of (y) = 6

    i was thinking of using the "let u =" thing but im not sure if that would work...


    6.) cube root of (x + 1) = sixth root of (4x + 9)


    i know theres quite a few but help on any of them would be awesome <33
    Last edited by eah1010; Feb 18th 2008 at 03:54 PM. Reason: wrong fraction in exponent
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eah1010 View Post
    1.) square root of (x + 6) - square root of (x-5) = 1
    $\displaystyle \sqrt{x + 6} - \sqrt{x - 5} = 1$

    Isolate one of the square roots.
    $\displaystyle \sqrt{x + 6} = \sqrt{x - 5} + 1$ <-- Square both sides

    $\displaystyle x + 6 = \left ( \sqrt{x - 5} + 1 \right ) ^2$

    $\displaystyle x + 6 = (x - 5) + 2 \sqrt{x - 5} + 1$ <-- Isolate the remaining square root

    $\displaystyle 2 \sqrt{x - 5} = 10$

    $\displaystyle \sqrt{x - 5} = 5$ <-- Square again

    $\displaystyle x - 5 = 25$

    $\displaystyle x = 30$

    Always always always check to see that this solution works in the original problem statement:
    $\displaystyle \sqrt{30 + 6} - \sqrt{30 - 5} = 1$ (Check!)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eah1010 View Post
    3.) x^2/3 + x^1/3 - 2 = 0
    This is a quadratic equation. Let $\displaystyle y = x^{1/3}$. Then $\displaystyle y^2 = x^{2/3}$, so the original equation is
    $\displaystyle y^2 + y - 2 = 0$

    Any way you like to solve it, we get
    $\displaystyle y = -2$ or $\displaystyle y = 1$

    Thus
    $\displaystyle x^{1/3} = -2$ or $\displaystyle x^{1/3} = 1$

    Cubing both sides gives
    $\displaystyle x = -8$ or $\displaystyle x = 1$

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by eah1010 View Post
    6.) cube root of (x + 1) = sixth root of (4x + 9)
    $\displaystyle \sqrt[3]{x + 1} = \sqrt[6]{4x + 9}$

    For clarity express the equation this way:
    $\displaystyle (x + 1)^{1/3} = (4x + 9)^{1/6}$

    Cube both sides. The exponent on the left becomes
    $\displaystyle \frac{1}{3} \cdot 3 = 1$

    and the exponent on the right becomes
    $\displaystyle \frac{1}{6} \cdot 3 = \frac{1}{2}$

    So
    $\displaystyle x + 1 = (4x + 9)^{1/2}$

    or
    $\displaystyle x + 1 = \sqrt{4x + 9}$

    I'm sure you're sick of doing this kind of problem, but you know how to continue from here.

    -Dan
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  5. #5
    Junior Member eah1010's Avatar
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    (:

    dan you're awesome thanks so much
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