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Math Help - Find the real solutions of the equation help

  1. #1
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    Find the real solutions of the equation help

    Can someone please tell me what I am doing wrong in the following calculation:

    x^4 + sqrt(3)*x^2 - 3 = 0
    sqrt(3)*x^2 = 3 - x^4 // I square all terms to get rid of the root
    3x^4 = 9 - x^8 // u = x^4
    3u = 9 - u^2
    -u^2 - 3u + 9

    //Using the quadratic formula to solve this
    [-(-3) +- sqrt((-3)^2 - 4(-1)(9))] / -2
    [3 +- sqrt(9 + 36)] / -2
    [3 +- sqrt(45)]/-2
    (3 +- 6.71) / -2
    u = (3 + 6.71) / -2 = -4.86
    u = (3 - 6.71)/-2 = 1.86

    The solution is supposed to be 1.03. I have tried to solve this several ways but somewhere I am going wrong. Any advice is really appreciated. Thanks in advance
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  2. #2
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    Re: Find the real solutions of the equation help

    Quote Originally Posted by alex95 View Post
    Can someone please tell me what I am doing wrong in the following calculation:

    x^4 + sqrt(3)*x^2 - 3 = 0
    sqrt(3)*x^2 = 3 - x^4 // I square all terms to get rid of the root
    3x^4 = 9 - x^8 // u = x^4
    3u = 9 - u^2
    -u^2 - 3u + 9

    //Using the quadratic formula to solve this
    [-(-3) +- sqrt((-3)^2 - 4(-1)(9))] / -2
    [3 +- sqrt(9 + 36)] / -2
    [3 +- sqrt(45)]/-2
    (3 +- 6.71) / -2
    u = (3 + 6.71) / -2 = -4.86
    u = (3 - 6.71)/-2 = 1.86

    The solution is supposed to be 1.03. I have tried to solve this several ways but somewhere I am going wrong. Any advice is really appreciated. Thanks in advance
    Start by letting \displaystyle \begin{align*} X = x^2 \end{align*} so that your equation becomes

    \displaystyle \begin{align*} X^2 + \sqrt{3}\,X - 3 &= 0 \\ X^2 + \sqrt{3}\,X &= 3 \\ X^2 + \sqrt{3}\,X + \left(\frac{\sqrt{3}}{2}\right)^2 &= 3 + \left(\frac{\sqrt{3}}{2}\right)^2 \\ \left( X + \frac{\sqrt{3}}{2} \right)^2 &= 3 + \frac{3}{4} \\ \left( X + \frac{\sqrt{3}}{2} \right)^2 &= \frac{15}{4} \\ X + \frac{\sqrt{3}}{2} &= \pm \frac{\sqrt{15}}{2} \\ X &= \frac{-\sqrt{3} \pm \sqrt{15}}{2} \\ x^2 &= \frac{-\sqrt{3} \pm \sqrt{15}}{2} \\ x &= \pm \sqrt{\frac{-\sqrt{3} \pm \sqrt{15}}{2}} \\ x &= \pm \sqrt{\frac{-\sqrt{3} + \sqrt{15}}{2}} \textrm{ since the other solution is nonreal} \\ x &= \pm \frac{\sqrt{2}\left(-\sqrt{3} + \sqrt{15} \right)}{2} \\ &= \pm \frac{\sqrt{30} - \sqrt{6}}{2} \end{align*}
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  3. #3
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    Re: Find the real solutions of the equation help

    Hello, alex95!

    x^4 + \sqrt{3}x^2 - 3 \:=\: 0

    \sqrt{3}x^2 \:=\: 3 - x^4

    I squared all terms to get rid of the root. . This is illegal!

    3x^4 \:=\: 9 - x^8 . This is wrong.

    You must square "legally":

    . . \begin{array}{c}\left(\sqrt{3}x^2\right)^2 \:=\:(3-x^4)^2 \\ \\[-3mm] 9x^4 \;=\;9 - 2x^4 + x^8  \\ \\[-3mm] x^8 - 11x^4 + 9 \;=\;0 \end{array}

    Now you can apply the Quadratic Formula.


    I applied the Quadratic first.

    . . x^4 + \sqrt{3}x^2 - 3 \;=\;0

    . . x^2 \;=\;\frac{\text{-}\sqrt{3} \pm\sqrt{(\sqrt{3})^2 - 4(1)(\text{-}3)}}{2(1)} \;=\;\frac{\text{-}\sqrt{3}\pm\sqrt{15}}{2}


    Discarding the negative root, we have:

    . . x \;=\;\pm\sqrt{\frac{\sqrt{15}-\sqrt{3}}{2}} \;=\;\pm1.034633399

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  4. #4
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    Re: Find the real solutions of the equation help

    Quote Originally Posted by Soroban View Post
    Hello, alex95!


    You must square "legally":

    . . 3-x^4)^2 \\ \\[-3mm] 9x^4 \;=\;9 - 2x^4 + x^8 \\ \\[-3mm] x^8 - 11x^4 + 9 \;=\;0 \end{array}" alt="\begin{array}{c}\left(\sqrt{3}x^2\right)^2 \:=\3-x^4)^2 \\ \\[-3mm] 9x^4 \;=\;9 - 2x^4 + x^8 \\ \\[-3mm] x^8 - 11x^4 + 9 \;=\;0 \end{array}" />

    Now you can apply the Quadratic Formula.


    I applied the Quadratic first.

    . . x^4 + \sqrt{3}x^2 - 3 \;=\;0

    . . x^2 \;=\;\frac{\text{-}\sqrt{3} \pm\sqrt{(\sqrt{3})^2 - 4(1)(\text{-}3)}}{2(1)} \;=\;\frac{\text{-}\sqrt{3}\pm\sqrt{15}}{2}


    Discarding the negative root, we have:

    . . x \;=\;\pm\sqrt{\frac{\sqrt{15}-\sqrt{3}}{2}} \;=\;\pm1.034633399

    You don't need to bother squaring both sides at all, it's ALREADY a quadratic...
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