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Math Help - Tough algebra

  1. #1
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    Tough algebra

    Solve for x in  \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0
    Last edited by Punch; February 11th 2010 at 08:17 PM.
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  2. #2
    Super Member Bacterius's Avatar
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    OMG, did you try solving a cubic polynomial using the quadratic formula or what ?

    Anyway, here's a start to simplification (I can't really do anything at the moment) : note that \frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}. Apply this to your equation with a = x^2 + 3
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  3. #3
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    I didn't. It's just an equation which I got from differentiation and am required to solve it...
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  4. #4
    Super Member Bacterius's Avatar
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    That was a mean curve, eh ...
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  5. #5
    Junior Member tedii's Avatar
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    \frac{\sqrt{x^2+3}}{2\sqrt{1-x}}=\frac{x\sqrt{1-x}}{\sqrt{x^2+3}}

    Since the top only needs to equal zero solve for this equation. This will become a lot easier when you get the denominators to match.
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  6. #6
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    Quote Originally Posted by tedii View Post
    \frac{\sqrt{x^2+3}}{2\sqrt{1-x}}=\frac{x\sqrt{1-x}}{\sqrt{x^2+3}}

    Since the top only needs to equal zero solve for this equation. This will become a lot easier when you get the denominators to match.
    Thanks, I think i got it, will try it now!
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  7. #7
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    this was what I did,  x^2+3=2x(1-x)

    x^2+3=2x-2x^2

    3x^2-2x+3=0

    Unable to solve further...
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  8. #8
    Junior Member tedii's Avatar
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    Quote Originally Posted by Punch View Post
    Solve for x in  \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0

    Looking at your derivative you made a mistake, I'm pretty sure. It should be

     \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{2x}{\sqrt{x^2+3}})}{x^2+3}=0

    Try redoing everything with that.
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  9. #9
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    Quote Originally Posted by Punch View Post
    Thanks, I think i got it, will try it now!
    Quote Originally Posted by tedii View Post
    Looking at your derivative you made a mistake, I'm pretty sure. It should be

     \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{2x}{\sqrt{x^2+3}})}{x^2+3}=0

    Try redoing everything with that.
    No I have checked and indeed I am right
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  10. #10
    Junior Member tedii's Avatar
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    Quote Originally Posted by Punch View Post
    Solve for x in  \frac{\sqrt{x^2+3}(\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0

    You're right about the 2 not being right.

     \frac{\sqrt{x^2+3}(-\frac{1}{2\sqrt{1-x}})-\sqrt{1-x}(\frac{x}{\sqrt{x^2+3}})}{x^2+3}=0

    But maybe you forgot the chain-rule on the first part for the -x.

    If not and you're sure your derivative is correct then your first equation has no maximums or minimums. And you don't need to find any zeros.
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  11. #11
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    Hello, Punch!

    \text{Solve for }x\!: \;\;\frac{\sqrt{x^2+3}\left(\dfrac{1}{2\sqrt{1-x}}\right)-\sqrt{1-x}\left(\dfrac{x}{\sqrt{x^2+3}}\right)}{x^2+3}\;\;  =\;\;0
    If this is a derivative, you lost a minus-sign . . .


    I assume the original function is: / f(x)\;=\;\frac{\sqrt{1-x}}{\sqrt{x^2+3}} \;=\;\frac{(1-x)^{\frac{1}{2}}} {(x^2+3)^{\frac{1}{2}}}


    Then: . f'(x) \;=\;\frac{(x^2+3)^{\frac{1}{2}}\cdot\frac{1}{2}(1-x)^{-\frac{1}{2}}{\color{red}(-1)} - (1-x)^{\frac{1}{2}}\cdot\frac{1}{2}(x^2+3)^{-\frac{1}{2}}(2x)}{x^2+3}

    . . . . . . . . =\;\frac{\sqrt{x^2+3}\left(\dfrac{{\color{red}-1}}{2\sqrt{1-x}}\right) - \sqrt{1-x}\left(\dfrac{x}{\sqrt{x^2+3}}\right)} {x^2+3}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Multiply top and bottom by 2\sqrt{x^2+3}\,\sqrt{1-x}

    . . and we have: . \frac{-(x^2+3) - 2x(1-x)}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0

    . . which simplifies to: . \frac{x^2-2x-3}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0


    And we have: . x^2 - 2x - 3 \:=\:0 \quad\Rightarrow\quad (x+1)(x-3) \:=\:0

    . . Hence: . x \;=\;-1,\;3


    But because of the \sqrt{1-x} in the function, x \leq 1


    . . Therefore, the only critical value is: . x \;=\;-1

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  12. #12
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    Quote Originally Posted by Soroban View Post
    Hello, Punch!

    If this is a derivative, you lost a minus-sign . . .


    I assume the original function is: / f(x)\;=\;\frac{\sqrt{1-x}}{\sqrt{x^2+3}} \;=\;\frac{(1-x)^{\frac{1}{2}}} {(x^2+3)^{\frac{1}{2}}}


    Then: . f'(x) \;=\;\frac{(x^2+3)^{\frac{1}{2}}\cdot\frac{1}{2}(1-x)^{-\frac{1}{2}}{\color{red}(-1)} - (1-x)^{\frac{1}{2}}\cdot\frac{1}{2}(x^2+3)^{-\frac{1}{2}}(2x)}{x^2+3}

    . . . . . . . . =\;\frac{\sqrt{x^2+3}\left(\dfrac{{\color{red}-1}}{2\sqrt{1-x}}\right) - \sqrt{1-x}\left(\dfrac{x}{\sqrt{x^2+3}}\right)} {x^2+3}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Multiply top and bottom by 2\sqrt{x^2+3}\,\sqrt{1-x}

    . . and we have: . \frac{-(x^2+3) - 2x(1-x)}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0

    . . which simplifies to: . \frac{x^2-2x-3}{2(x^2+3)^{\frac{3}{2}}\sqrt{1-x}} \;=\;0


    And we have: . x^2 - 2x - 3 \:=\:0 \quad\Rightarrow\quad (x+1)(x-3) \:=\:0

    . . Hence: . x \;=\;-1,\;3


    But because of the \sqrt{1-x} in the function, x \leq 1


    . . Therefore, the only critical value is: . x \;=\;-1
    Sorry Soroban, this is just how careless I am... thanks
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