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Math Help - surds

  1. #1
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    surds

    find the number of solutions in (5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10
    Last edited by sciencepal; May 1st 2013 at 06:42 AM.
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  2. #2
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    Re: surds

    Quote Originally Posted by sciencepal View Post
    find the number of solutions in (5 + 2*6^0.5)^(x^2-3) + (5 - 2*6^0.5)^(x^2-3) = 10
    Why don’t you take some time to learn to post symbols?
    It really is so easy. And it makes most of us more willing to find out how to help.
    This subforum will help you with the code. Once you begin, you quickly learn the code.
    LaTeX Help

    [TEX](5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10 [/TEX] gives  (5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10
    If you click on the “go advanced tab” you should see \boxed{\Sigma} on the tool-bar. That gives the [TEX]..[/TEX] wrap. Your LaTeX code goes between them.

    Now it should be perfectly clear the two solutions are x=\pm 2~.
    Thanks from topsquark
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  3. #3
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    Re: surds

    Thanks for your advice but there are 4 solutions

    \pm\sqrt{2}
    \pm2
    Could you please solve the equation?
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  4. #4
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    Re: surds

    Quote Originally Posted by sciencepal View Post
    Thanks for your advice but there are 4 solutions

    \pm\sqrt{2}
    \pm2
    Could you please solve the equation?
    Show that \pm\sqrt2 are solutions.

    What does {\left( {5 + 2\sqrt 6 } \right)^{ - 1}} = ~?
    Thanks from topsquark
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  5. #5
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    Re: surds

    That's fine but how do I exactly know how many solutions can the equation have?
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    Re: surds

    Quote Originally Posted by sciencepal View Post
    That's fine but how do I exactly know how many solutions can the equation have?
    As far as I know there is no elementary solution to this question.

    Drawing a graph would give you some idea.
    Thanks from sciencepal
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  7. #7
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    Re: surds

    Well now I got the idea. This equation holds true when the powers are 1 or -1 and a^{2}-b=1. In this case 5^{2}-(2\sqrt{6})^{2}=1. Thanks.
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  8. #8
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    Re: surds

    Hello, sciencepal!

    \text{Find the number of solutions in: }\:(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}\:=\:10

    We note that: . 5\pm2\sqrt{6} \:=\:\left(\sqrt{3}\pm\sqrt{2}\right)^2

    The equation becomes: . \left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} + \left(\sqrt{3}-\sqrt{2}\right)^{2(x^2-3)} \:=\:10

    We further note that: . \sqrt{3}-\sqrt{2} \:=\:\frac{\sqrt{3}-\sqrt{2}}{1}\cdot\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}  +\sqrt{2}} \:=\: \frac{1}{\sqrt{3}+\sqrt{2}}

    The equation becomes: . \left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} + \frac{1}{(\sqrt{3}+\sqrt{2})^{2(x^2-3)}} \:=\:1


    Multiply by \left(\sqrt{3}+\sqrt{3}\right)^{2(x^2-3)}

    . . \left(\sqrt{3}+\sqrt{2}\right)^{4(x^2-3)} + 1 \:=\:10\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)}

    . . \left(\sqrt{3}+\sqrt{2}\right)^{4(x^2-3)} - 10\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} + 1 \;=\;0


    Let u \:=\:\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-30}

    We have: . u^2 - 10u + 1 \:=\:0

    Quadratic Formula: . u \;=\;5 \pm2\sqrt{3} \;=\;\left(\sqrt{3}\pm\sqrt{2}\right)^2

    Back-substitute: . \left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} \;=\;\left(\sqrt{3}\pm\sqrt{2}\right)^2

    Take logs: . \ln\left(\sqrt{3}+\sqrt{2}\right)^{2(x^2-3)} \;=\;\ln\left(\sqrt{3}\pm\sqrt{2}\right)^2

    . . 2(x^2-3)\ln\left(\sqrt{3}+\sqrt{2}\right) \;=\;2\ln\left(\sqrt{3}\pm\sqrt{2}\right)

    . . x^2-3 \;=\;\frac{\ln(\sqrt{3}\pm\sqrt{2})}{\ln(\sqrt{3}+  \sqrt{2})}


    The \pm gives us 2 solutions.

    And the x^2 gives us 2\cdot 2 \:=\:4 solutions.
    Thanks from agentmulder
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