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Math Help - Solve equation

  1. #1
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    Solve equation

    Solve equation-capture.jpg
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  2. #2
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    Re: Solve equation

    Quote Originally Posted by Mhmh96 View Post
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    Question in Latex: x = 12-\sqrt{12-\sqrt{x}}

    What have you tried? A good start would be to rearrange the equation to isolate the radical term so it's easier to square both sides

    From the domain we know that x \geq 0 \text{  and  } \sqrt{x} \leq 12 \Leftrightarrow x \leq 144. Overall this means 0 \leq x \leq 144

    Edit: Don't forget to test potential solutions in the original equation to check if they are spurious or not
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  3. #3
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    Re: Solve equation

    I tried to square both side but still not able to
    determine to the right answer which is 9
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  4. #4
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    Re: Solve equation

    Hello, Mhmh96!

    \text{Solve for }x\!:\;\;x \;=\;12 - \sqrt{12-\sqrt{x}}

    Let u \,=\, \sqrt{x}\quad\Rightarrow\quad x \,=\,u^2

    We have: . u^2 \:=\:12 - \sqrt{12-u} \quad\Rightarrow\quad \sqrt{12-u} \;=\;12 - u^2

    Square both sides: . \left(\sqrt{12-u}\right)^2 \;=\; \left(12-u^2\right)^2

    . . 12-u \;=\;144 - 24u^2 + u^4 \quad\Rightarrow\quad u^4 - 24u^2 + u + 132 \:=\:0

    which factors: . (u-3)(u+4)(u^2-u-11) \:=\:0


    We have: . \begin{Bmatrix} u-3 \:=\:0 & \Rightarrow & u \:=\:3 \\ \\[-3mm] u+4\:=\:0 & \Rightarrow & u \:=\:\text{-}4 \\ u^2-u-11\:=\:0 & \Rightarrow & u \:=\:\dfrac{1\pm3\sqrt{5}}{2} \end{Bmatrix}

    Back-substitute and solve for x.
    . . Be sure to check for extraneous roots.

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