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Math Help - Polynomial equation

  1. #1
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    Polynomial equation

    (x+2)/2=2√(x+1)

    Have tried a lot of things, but I am pretty certain that I am making a mistake in my initial steps so I dont think
    a copy of my tries would be of help.

    thanks in advance
    Last edited by Ippo; April 9th 2013 at 03:16 PM.
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  2. #2
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    Re: Polynomial equation

    Have you tried denoting x^2 + 1 by t?
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  3. #3
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    Re: Polynomial equation

    not specifically x^2 + 1 , but i have tried the method , and after reducing the equation to x^4 - 16x^3 - 12 = 0 - I am blank .
    Any way to get a Real number solution ?
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  4. #4
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    Re: Polynomial equation

    First, that's not a polynomial equation. Second there is a missing "(" on the left. Is it supposed to be (x^2+ 2)/2?
    And it is hard to read- that is x^3+ 1 on the right, not x^2+ 1 as emkarov seemed to think.

    The first thing you should do is make it a polynomial equation by squaring both sides.
    \frac{x^4+ 4x^2+ 4}{4}= 4(x^3+ 1)
    x^4+ 4x^2+ 4= 16x^3+ 16
    x^4- 16x^3+ 4x^2- 12= 0.

    By the "rational root theorem" the only possible roots of that are the factor of 12: 1, -1, 2, -2, 4, -4, 8, -8, 12, and -12.

    And unless I have done the arithemtic wrong none of those satisfy the equation so the equation has only irrational or complex roots.

    Quote Originally Posted by Ippo View Post
    x+2)/2=2√(x+1)

    Have tried a lot of things, but I am pretty certain that I am making a mistake in my initial steps so I dont think
    a copy of my tries would be of help.

    thanks in advance
    Last edited by HallsofIvy; April 9th 2013 at 03:21 PM.
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  5. #5
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    Re: Polynomial equation

    what happened to the 4 in the denominator
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  6. #6
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    Re: Polynomial equation

    I knew some one would catch me! It took me too long to correct it.
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  7. #7
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    Re: Polynomial equation

     \text{Currently I cannot find a substitution method to solve this problem. However, it can be solved by brute force!}

     x^2+2=4\sqrt{x^3+1}~\Leftrightarrow~x^4-16x^3+4x^2-12=0\qquad (1)

     \text{There is no real rational solution to (1), therefore we have to solve the \emph{quartic equation} using \textbf{Ferrari's Method}}.

     \text{Let}~ x=y-\frac{b}{4a}~\Rightarrow~x=y+4.~\text{This substitution will transform}~(1)~\text{into:}

     y^4-92y^2-480y-716=0~\Leftrightarrow~y^4=92y^2+480y+716

     \text{Add}~2zy^2+z^2~\text{to both sides of the equation, for some}~z\in R.~\text{The added terms will make the left side of the equation a perfect square.}

     \left(y^2+z\right)^2=(2z+92)y^2+480y+716+z^2\qquad (2)

     \text{The right side is a perfect square if its discriminant is equal to 0, since it is a quadratic equation in terms of}~y.

     D=230400-4(2z+92)(z^2+716)=0~\Leftrightarrow~28800-(z+46)(z^2+716)=0

     \Leftrightarrow~z^3+46z^3+716z+4136=0\qquad(3)

     \text{Let}~z=u-\frac{b}{3a}~\Leftrightarrow~z=u-\frac{46}{3}\qquad (4)

     (3)~\Leftrightarrow~u^3+\frac{32}{3}u+\frac{9920}{  27}~\Leftrightarrow~27u^3+288u+9920=0

     \Leftrightarrow~(3u+20)(9u^2-60u+496)=0~\Rightarrow~u=-\frac{20}{3}~\Rightarrow~z=-22

     (2)~\Leftrightarrow~\left(y^2-22\right)^2=48y^2+480y+1200=48(y+5)^2

     \Leftrightarrow~y^2-22=\pm4\sqrt{3}(y+5)~\Rightarrow~\begin{cases} y^2-4\sqrt{3}y-\left(22+20\sqrt{3}\right)=0\qquad (5)\\  y^2+4\sqrt{3}y-\left(22-20\sqrt{3}\right)=0\qquad (6) \end{cases}

     (5)~\Rightarrow~y=2\sqrt{3}\pm\sqrt{34+20\sqrt{3}}  ~\Rightarrow~x=4+2\sqrt{3}\pm\sqrt{34+20\sqrt{3}}

     (6)~\text{has no real solution.}

     \text{Thus the original equation has two real solutions.}

     x_1=4+2\sqrt{3}+\sqrt{34+20\sqrt{3}}\simeq15.7491

     \text{and}

     x_2=4+2\sqrt{3}-\sqrt{34+20\sqrt{3}}\simeq-0.8208
    Last edited by thevinh; September 5th 2013 at 07:48 PM.
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