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Thread: Find All Real Solutions

  1. September 13th 2010, 08:14 AM #1
    fdrhs1984
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    Find All Real Solutions

    Find all real solutions of the equation exactly.

    10x^4 + 3x^2 = 1
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  2. September 13th 2010, 08:16 AM #2
    undefined
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    Quote Originally Posted by fdrhs1984 View Post
    Find all real solutions of the equation exactly.

    10x^4 + 3x^2 = 1
    What have you tried?

    Rule # 11, show some effort.

    http://www.mathhelpforum.com/math-he...ng-151421.html

    That said, if you substitute y = x^2 you will get a quadratic equation.
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  3. September 13th 2010, 08:33 AM #3
    Soroban
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    Hello, fdrhs1984!

    Did you try anything?
    Are you waiting for Divine Intervention?

    Find all real solutions of the equation: . 10x^4 + 3x^2 \:=\:1

    We have: . 10x^4 + 3x^2 - 1 \:=\:0

    Factor: . (2x^2+1)(5x^2-1) \:=\:0


    And we have: . \begin{Bmatrix}2x^2 + 1 \:=\:0 & \Rightarrow & x^2 \:=\:-\frac{1}{2} && \text{no real roots} \\ \\[-3mm]<br /> 5x^2 - 1 \:=\:0 & \Rightarrow & x^2 \:=\:\frac{1}{5} & \Rightarrow & x \:=\:\pm\frac{1}{\sqrt{5}} \end{Bmatrix}
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  4. September 14th 2010, 05:27 AM #4
    fdrhs1984
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    Quote Originally Posted by Soroban View Post
    Hello, fdrhs1984!

    Did you try anything?
    Are you waiting for Divine Intervention?


    We have: . 10x^4 + 3x^2 - 1 \:=\:0

    Factor: . (2x^2+1)(5x^2-1) \:=\:0


    And we have: . \begin{Bmatrix}2x^2 + 1 \:=\:0 & \Rightarrow & x^2 \:=\:-\frac{1}{2} && \text{no real roots} \\ \\[-3mm]<br /> 5x^2 - 1 \:=\:0 & \Rightarrow & x^2 \:=\:\frac{1}{5} & \Rightarrow & x \:=\:\pm\frac{1}{\sqrt{5}} \end{Bmatrix}

    I don't understand how you got (2x^2 + 1)(5x^2 - 1).
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  5. September 14th 2010, 06:02 AM #5
    CaptainBlack
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    Quote Originally Posted by fdrhs1984 View Post
    I don't understand how you got (2x^2 + 1)(5x^2 - 1).
    Put y=x^2 then you have a quadratic in $$ y to solve, from the solutions you can then find the $$ x's that solve the original equation.

    CB
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