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Math Help - Need Help With This 1 Problem Please...

  1. #1
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    Need Help With This 1 Problem Please...



    okay so our teacher gave us this problem today and im a little confused...i have answers but i wanted to see if anyone got the same ones so i'd know if i was anywhere close to being right...any help would be appreciated...

    <edit>

    my teacher said to solve for x
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  2. #2
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    If the question is to simplify the equation, you need to square it once, then rearrange the equation so that the remaining square root is on one side of the equation, and then square the equation again.
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  3. #3
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    Hello, hc0company!

    \sqrt{x+a^2} - \sqrt{x-2a^2} \;=\;\sqrt{2x-5a^2}

    Square both sides: . \left(\sqrt{x+a^2} - \sqrt{x-2a^2}\right)^2 \;=\;\left(\sqrt{2x-5a^2}\right)^2

    . . x + a^2 - 2\sqrt{(x+a^2)(x-2a^2)} + x - 2a^2 \;=\;2x-5a^2

    . . 2\sqrt{(x+a^2)(x - 2a^2)} \;=\;4a^2 \quad\Rightarrow\quad \sqrt{(x+a^2)(x-2a^2)} \;=\;2a^2


    Square both sides: . \left(\sqrt{(x+a^2)(x-2a^2)}\right)^2 \;=\;\left(2a^2\right)^2

    . . (x+a^2)(x-2a^2) \;=\;4a^4 \quad\Rightarrow\quad x^2 - a^2x - 6a^4 \;=\;0


    Factor: . (x - 3a^2)(x + 2a^2) \;=\;0


    Therefore: . \begin{array}{cccc}x - 3a^2 \:=\:0 & \Rightarrow & \boxed{x \:=\:3a^2} \\ \\[-4mm]<br />
x + 2a^2 \:=\: 0 & \Rightarrow & {\color{red}\rlap{/////////}}x \:=\:-2a^2 & \text{extraneous root} \end{array}

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, hc0company!


    Square both sides: . \left(\sqrt{x+a^2} - \sqrt{x-2a^2}\right)^2 \;=\;\left(\sqrt{2x-5a^2}\right)^2

    . . x + a^2 - 2\sqrt{(x+a^2)(x-2a^2)} + x - 2a^2 \;=\;2x-5a^2

    . . 2\sqrt{(x+a^2)(x - 2a^2)} \;=\;4a^2 \quad\Rightarrow\quad \sqrt{(x+a^2)(x-2a^2)} \;=\;2a^2


    Square both sides: . \left(\sqrt{(x+a^2)(x-2a^2)}\right)^2 \;=\;\left(2a^2\right)^2

    . . (x+a^2)(x-2a^2) \;=\;4a^4 \quad\Rightarrow\quad x^2 - a^2x - 6a^4 \;=\;0


    Factor: . (x - 3a^2)(x + 2a^2) \;=\;0


    Therefore: . \begin{array}{cccc}x - 3a^2 \:=\:0 & \Rightarrow & \boxed{x \:=\:3a^2} \\ \\[-4mm]<br />
x + 2a^2 \:=\: 0 & \Rightarrow & {\color{red}\rlap{/////////}}x \:=\:-2a^2 & \text{extraneous root} \end{array}

    the exact answers i got man...thanks for the help...
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