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Math Help - Radical Equations #1

  1. #1
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    Post Radical Equations #1

    http://puu.sh/3kEIh.png

    My attempt: I wrote it down on paper and I have nothing I could use to take a picture of, I'm not sure how I can write it out but:

    square both sides to get rid of the radical sign on the left side then use FOIL on right side.

    q^2/2 + 11 = q^2 - 2q + 2

    multiply 2

    q^2 - 4q - 18

    then i put it into the quadratic formula, q^2 = a -4 = b -18 = c

    after that, I ended up with http://puu.sh/3kEO7.png

    the answer for this question is suppose to be q = 2 + 2(sqrt(6))

    i don't know how to get the right answer.
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  2. #2
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    Re: Radical Equations #1

    On checking we find tat for this equation we will not get the answer given. please check if the question has been correctly pasted.
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  3. #3
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    Re: Radical Equations #1

    yes, i pasted my question correctly.
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  4. #4
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    Re: Radical Equations #1

    Radical Equations #1-21-jun-13.png
    Thanks from topsquark
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  5. #5
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    Re: Radical Equations #1



    this question solved is suppose to be q = 2 + 2(sqrt(6)) {apparently that is the correct answer}

    but I get http://puu.sh/3kEO7.png
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  6. #6
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    Re: Radical Equations #1

    Hello, Mathnood768!

    You're making a number of blunders . . .


    \sqrt{\frac{q^2}{2} + 11} \:=\:q-1

    Square both sides: . \left(\sqrt{\frac{q^2}{2} + 11}}\right)^2 \;=\;(q-1)^2

    . . . . . . . . . . . . . . . . . . . \frac{q^2}{2} + 11 \;=\;q^2-2q + 1

    Multiply by 2: . . . . . . . . . q^2 + 22 \;=\;2q^2 - 4q + 2

    . . . . . . . . . . . . . . . . q^2 - 4q - 20 \;=\;0


    Quadratic Formula: . q \;=\;\frac{\text{-}(\text{-}4) \pm\sqrt{(\text{-}4)^2 - 4(1)(\text{-}20)}}{2(1)} \;=\;\frac{4\pm\sqrt{96}}{2}

    . . . . . . . . . . . . . . . q \;=\;\frac{4\pm 4\sqrt{6}}{2} \;=\;\frac{2(2\pm2\sqrt{6})}{2} \;=\;2 \pm2\sqrt{6}


    But we find that. q \:=\:2-2\sqrt{6}ndoes not check.
    . . It is an extraneous root.

    The only root is: . q \:=\:2 + 2\sqrt{6}
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