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Math Help - Challenge Algebra Equations

  1. #1
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    Challenge Algebra Equations

     \text{Solve for \emph{x}. Real solutions only.}

     64x^6-112x^4+56x^2-7=2\sqrt{1-x^2}

     x^5-15x^3-45x-27=0
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  2. #2
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    Re: Challenge Algebra Equations

    Wait, are the two equations separate questions or does x have to satisfy both equations?
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  3. #3
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    Re: Challenge Algebra Equations

    Quote Originally Posted by Mrdavid445 View Post
    Wait, are the two equations separate questions or does x have to satisfy both equations?
    They are two separate equations. Solve for x in each equation. Have fun!
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  4. #4
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    Re: Challenge Algebra Equations

    Quote Originally Posted by thevinh View Post
     \text{Solve for \emph{x}. Real solutions only.}
     x^5-15x^3-45x-27=0
    x = 4.22786... or -4.15136... or -.54664... ; other 2 are imaginary.

    Change the -27 to +116, then x = 4 exactly.

    Any purpose to this madness ?
    Last edited by Wilmer; August 8th 2012 at 06:58 PM.
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  5. #5
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    Re: Challenge Algebra Equations

    Wolfram Alpha ftw.
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  6. #6
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    Re: Challenge Algebra Equations

    Quote Originally Posted by Wilmer View Post
    x = 4.22786... or -4.15136... or -.54664... ; other 2 are imaginary.

    Change the -27 to +116, then x = 4 exactly.

    Any purpose to this madness ?
     \text{Hi, I realized I made a mistake in posting the equation. I am so sorry about that. The solution is posted below.}

     \textbf{Problem 2:}\quad \text{Solve}\mspace{7mu} x^5-15x^3+45x-27=0 \quad\quad (1)

     \text{From the Rational Root Theorem,}\mspace{5mu} x=-3,\mspace{5mu} \text{is one of the many solution}.

     \text{Thus}\mspace{7mu} (1)\, \Leftrightarrow\, (x+3)(x^4-3x^3-6x^2+18x-9)=0

     x^4-3x^3-6x^2+18x-9=0\, \Leftrightarrow\, x^4-3x^2(x-1)-9(x-1)^2=0

     \text{Let}\mspace{7mu} y=x-1\, \Rightarrow\, x^4-3yx^2-9y^2=0\quad \quad (2)

     \text{Solve for \emph{x} in terms of \emph{y}, we get:}\mspace{7mu} 2x^2=3y\pm3y\sqrt{5}

     \text{\underline{Case 1.}}\quad 2x^2=3y+3y\sqrt{5}\, \Leftrightarrow\, 2x^2=3(x-1)+3\sqrt{5}(x-1)

     \Leftrightarrow\, 2x^2-3\left(\sqrt{5}+1\right)x+3\left(\sqrt{5}+1\right)  =0

     \text{Solve the quadratic equation, we get:}

     x=\dfrac{1}{4} \left(3+3\sqrt{5} \pm \sqrt{30-6\sqrt{5}}\mspace{1mu} \right)

     \text{\underline{Case 2.}}\quad 2x^2=3y-3y\sqrt{5}\, \Leftrightarrow\, 2x^2=3(x-1)-3\sqrt{5}(x-1)

     \Leftrightarrow\, 2x^2+3\left(\sqrt{5}-1\right)x-3\left(\sqrt{5}-1\right)=0

     \text{Solve the quadratic equation, we get:}

     x=\dfrac{1}{4} \left(3-3\sqrt{5} \pm \sqrt{30+6\sqrt{5}}\mspace{1mu} \right)

     \text{So (1) has five real solutions:}

     x=-3

     x=\dfrac{1}{4} \left(3+3\sqrt{5} \pm \sqrt{30-6\sqrt{5}}\mspace{1mu} \right)

     x=\dfrac{1}{4} \left(3-3\sqrt{5} \pm \sqrt{30+6\sqrt{5}}\mspace{1mu} \right)
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    Re: Challenge Algebra Equations

    Quote Originally Posted by thevinh View Post
     \text{Hi, I realized I made a mistake in posting the equation. I am so sorry about that. The solution is posted below.}

     \textbf{Problem 2:}\quad \text{Solve}\mspace{7mu} x^5-15x^3+45x-27=0 \quad\quad (1)

     \text{From the Rational Root Theorem,}\mspace{5mu} x=-3,\mspace{5mu} \text{is one of the many solution}.

     \text{Thus}\mspace{7mu} (1)\, \Leftrightarrow\, (x+3)(x^4-3x^3-6x^2+18x-9)=0

     x^4-3x^3-6x^2+18x-9=0\, \Leftrightarrow\, x^4-3x^2(x-1)-9(x-1)^2=0

     \text{Let}\mspace{7mu} y=x-1\, \Rightarrow\, x^4-3yx^2-9y^2=0\quad \quad (2)

     \text{Solve for \emph{x} in terms of \emph{y}, we get:}\mspace{7mu} 2x^2=3y\pm3y\sqrt{5}

     \text{\underline{Case 1.}}\quad 2x^2=3y+3y\sqrt{5}\, \Leftrightarrow\, 2x^2=3(x-1)+3\sqrt{5}(x-1)

     \Leftrightarrow\, 2x^2-3\left(\sqrt{5}+1\right)x+3\left(\sqrt{5}+1\right)  =0

     \text{Solve the quadratic equation, we get:}

     x=\dfrac{1}{4} \left(3+3\sqrt{5} \pm \sqrt{30-6\sqrt{5}}\mspace{1mu} \right)

     \text{\underline{Case 2.}}\quad 2x^2=3y-3y\sqrt{5}\, \Leftrightarrow\, 2x^2=3(x-1)-3\sqrt{5}(x-1)

     \Leftrightarrow\, 2x^2+3\left(\sqrt{5}-1\right)x-3\left(\sqrt{5}-1\right)=0

     \text{Solve the quadratic equation, we get:}

     x=\dfrac{1}{4} \left(3-3\sqrt{5} \pm \sqrt{30+6\sqrt{5}}\mspace{1mu} \right)

     \text{So (1) has five real solutions:}

     x=-3

     x=\dfrac{1}{4} \left(3+3\sqrt{5} \pm \sqrt{30-6\sqrt{5}}\mspace{1mu} \right)

     x=\dfrac{1}{4} \left(3-3\sqrt{5} \pm \sqrt{30+6\sqrt{5}}\mspace{1mu} \right)

    It's looking right and but to be sure we have to wait thread starter to see if he has another method to solve this.algebra word problem solver free
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