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Thread: Solve the equation

  1. #1
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    Solve the equation

    $\displaystyle \frac 1{\left(1+\sqrt{1+\sqrt{x}}\right)^4}+\frac 1{\left(1-\sqrt{1+\sqrt{x}}\right)^4}+\frac 2{\left(1+\sqrt{1+\sqrt{x}}\right)^3}+\frac 2{\left(1-\sqrt{1+\sqrt{x}}\right)^3}=0$
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by james_bond View Post
    $\displaystyle \frac 1{\left(1+\sqrt{1+\sqrt{x}}\right)^4}+\frac 1{\left(1-\sqrt{1+\sqrt{x}}\right)^4}+\frac 2{\left(1+\sqrt{1+\sqrt{x}}\right)^3}+\frac 2{\left(1-\sqrt{1+\sqrt{x}}\right)^3}=0$
    Put $\displaystyle y=1+\sqrt{x}$, then your equation becomes:

    $\displaystyle \frac 1{\left(1+\sqrt{y}\right)^4}+\frac 1{\left(1-\sqrt{y}\right)^4}+\frac 2{\left(1+\sqrt{y}\right)^3}+\frac 2{\left(1-\sqrt{y}\right)^3}=0$

    $\displaystyle \frac {\left(1-\sqrt{y}\right)^4} {(1-y)^4} +
    \frac {\left(1+\sqrt{y}\right)^4}{(1-y)^4}+
    \frac {2\left(1-\sqrt{y}\right)^3}{(1-y)^3}+
    \frac {2\left(1+\sqrt{y}\right)^3}{(1-y)^3}=0$

    and since $\displaystyle y \ge 0$:

    $\displaystyle \left(1-\sqrt{y}\right)^4 +
    \left(1+\sqrt{y}\right)^4+
    2\left(1-\sqrt{y}\right)^3(1-y)+
    2\left(1+\sqrt{y}\right)^3(1-y)=0$


    Now expand the powers and simplify and see what you have.

    RonL
    Last edited by CaptainBlack; Feb 5th 2008 at 07:34 PM.
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  3. #3
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    I posted a carefully crafted reply to this before it was moved.

    What happened to my post?

    I started with $\displaystyle a=\sqrt{1+\sqrt{x}}$
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    $\displaystyle \left(1-\sqrt{y}\right)^4 +
    \left(1+\sqrt{y}\right)^4+
    2\left(1-\sqrt{y}\right)^3(1+y)+
    2\left(1+\sqrt{y}\right)^3(1+y)=0$


    Now expand the powers and simplify and see what you have.
    I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
    From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
    But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).
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  5. #5
    GAMMA Mathematics
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    Quote Originally Posted by james_bond View Post
    I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
    From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
    But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).
    Let me check this out for you...

    $\displaystyle 14y^2+28y+6=0$

    $\displaystyle y^2+2y+\frac{3}{7}=0$

    $\displaystyle y=\frac{-2 \pm \sqrt{4-4\frac{3}{7}}}{2}$

    $\displaystyle y=\frac{-2 \pm \sqrt{\frac{28-12}{7}}}{2}$

    $\displaystyle y=\frac{-2 \pm \sqrt{\frac{16}{7}}}{2}$

    $\displaystyle y=\frac{-2 \pm 4\sqrt{\frac{1}{7}}}{2}$

    $\displaystyle y=-1 \pm 2\sqrt{\frac{1}{7}}$

    A solution of $\displaystyle y=1.6$ is impossible if
    $\displaystyle 14y^2+28y+6=0$ is the quadratic equation. Notice that any positive number makes the equation greater than zero, and never less than or equal to zero.
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  6. #6
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    Quote Originally Posted by colby2152 View Post

    A solution of $\displaystyle y=1.6$ is impossible if
    $\displaystyle 14y^2+28y+6=0$ is the quadratic equation. Notice that any positive number makes the equation greater than zero, and never less than or equal to zero.
    Yes but why would $\displaystyle y$ be $\displaystyle 1.6$
    Both roots are negative. Ohh so $\displaystyle \sqrt{x}$ would be negative - contradiction - no solutions.
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  7. #7
    GAMMA Mathematics
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    Quote Originally Posted by james_bond View Post
    Yes but why would $\displaystyle y$ be $\displaystyle 1.6$
    Both roots are negative. Ohh so $\displaystyle \sqrt{x}$ would be negative - contradiction - no solutions.
    The discriminant would not be negative. I used the quadratic formula in my solution. There are two solutions, and both are negative real numbers. (SIDE NOTE: I suppose this is a contradiction since to be negative implies real?)
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  8. #8
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    Quote Originally Posted by james_bond View Post
    I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
    From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
    But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).
    Now I think you should get: $\displaystyle 10y^2-20y-6=0$

    (which if I've got this right gives a root of $\displaystyle \approx 1.6$, which agrees with
    the graphical sdolution)

    (I must admit that I used "a tutors" substitution rather than my own as
    it is easier to simplify - and then convert back as if I had made my substitution)

    RonL
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by james_bond View Post
    I used Wolfram Mathematica and got: $\displaystyle 14y^2+28y+6=0$
    From this quadratic equation $\displaystyle y=\frac 17 \left(-7 \pm 2\sqrt{7}\right)\rightarrow x=\pm \frac 87 \left(-4 + \sqrt{7}\right)$
    But it is not a solution. I can't find any mistakes in your solution (and I did it quite similarly) but the only solution is $\displaystyle \frac 85$ (by Mathematica ).
    Sorry, but note the deliberate (or was it just a transcription) error, now
    corrected.

    RonL
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by a tutor View Post
    I posted a carefully crafted reply to this before it was moved.

    What happened to my post?

    I started with $\displaystyle a=\sqrt{1+\sqrt{x}}$
    Oppsss.. that will be me. Over enthusiatic moderating (we had a glut of
    multiple posts and other moderation issues yesterday)

    RonL
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