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Thread: Equation

  1. #1
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    Equation

    solve for x the equation :
    $\displaystyle x+ \sqrt{a^2+x^2} = \frac{5a^2}{ a^2+b^2}$
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by beq!x View Post
    solve for x the equation :
    $\displaystyle x+ \sqrt{a^2+x^2} = \frac{5a^2}{ a^2+b^2}$
    Take x from both sides and square:

    $\displaystyle a^2+x^2 = (\frac{5a^2}{ a^2+b^2} - x)^2 = \frac{25a^4}{(a^2+b^2)^2} - \frac{10a^2x}{a^2+b^2} + x^2$

    $\displaystyle x^2$ will cancel. Rearranging gives:

    $\displaystyle \frac{10a^2}{a^2+b^2}x = \frac{25a^4}{(a^2+b^2)^2}$

    From there solve for x
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  3. #3
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    Quote Originally Posted by beq!x View Post
    $\displaystyle x+ \sqrt{a^2+x^2} = \frac{5a^2}{ a^2+b^2}$
    here is a mistake it is not $\displaystyle b$ but $\displaystyle x$, sorry.
    the right equation: $\displaystyle x+ \sqrt{a^2+x^2} = \frac{5a^2}{ a^2+x^2}$

    i have a question for e^(i*pi) : did you forget $\displaystyle a^2$ here : $\displaystyle \frac{10a^2}{a^2+b^2}x = \frac{25a^4}{(a^2+b^2)^2}
    $ or am i wrong?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Subract x from both sides and then square both sides. I think that is how I would do it.
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  5. #5
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    root questions

    if $\displaystyle \sqrt{a^{2}+x^{2}}$ can we do this$\displaystyle \sqrt{a^2} + \sqrt{x^2}= a+ x$
    or what should i do ???
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  6. #6
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    Quote Originally Posted by beq!x View Post
    if $\displaystyle \sqrt{a^{2}+x^{2}}$ can we do this$\displaystyle \sqrt{a^2} + \sqrt{x^2}= a+ x$
    No, you can't!
    Last edited by mr fantastic; May 5th 2009 at 05:47 AM. Reason: Moved from another thread.
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  7. #7
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    Quote Originally Posted by beq!x View Post
    if $\displaystyle \sqrt{a^{2}+x^{2}}$ can we do this$\displaystyle \sqrt{a^2} + \sqrt{x^2}= a+ x$
    or what should i do ???
    Does your conjecture hold when a = 1 and x = 1? (This is called a counter-example).
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