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Math Help - Simplify

  1. #1
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    Simplify

    M = 2 - square_root(u - 3)
    u = (v - 2)^2 + 3

    Prove that M = v.
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  2. #2
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    Re: Simplify

    Quote Originally Posted by joshuaa View Post
    M = 2 - square_root(u - 3)
    u = (v - 2)^2 + 3

    Prove that M = v.
    $M = 2 - \sqrt{u - 3} \implies \sqrt{u - 3} = 2 - M.$ I wonder what might happen if you now square both sides?
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  3. #3
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    Re: Simplify

    I don't agree this is the convenient way. I prefer to do it this WAY.

    M = 2 - square_root((v - 2)^2 + 3 - 3) = 2 - v - 2 = - v

    Why does my answer become -v, Not +v?
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  4. #4
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    Re: Simplify

    Your use of the square root is incorrect. \sqrt{x^2}= |x|, NOT x.
    So \sqrt{(v- 2)^2}= |v- 2|. Whether that is equal to v- 2 or 2- v depends upon whether v\ge 2 or v< 2.

    If you are not told whether v is larger than 2 or not, the best you can say is that
    2- \sqrt{((v- 2)^2+ 3)- 3}= 2- \sqrt{(v- 2)^2}= 2- |v- 2|. That is either v (if v\ge 2) or 4- v (if v\le 2).
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  5. #5
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    Re: Simplify

    Quote Originally Posted by joshuaa View Post
    I don't agree this is the convenient way. I prefer to do it this WAY.

    M = 2 - square_root((v - 2)^2 + 3 - 3) = 2 - v - 2 = - v

    Why does my answer become -v, Not +v?
    $\sqrt{(v-2)^2+3-3}=|v-2|$ So I do not agree with the premise of the question.
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  6. #6
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    Re: Simplify

    Quote Originally Posted by joshuaa View Post
    I don't agree this is the convenient way. I prefer to do it this WAY.

    M = 2 - square_root((v - 2)^2 + 3 - 3) = 2 - v - 2 = - v

    Why does my answer become -v, Not +v?
    The way the square root is defined on real numbers

    $\sqrt{x^2}=|x|$

    so you end up with

    $M=2-|v-2|$

    $M=\begin{cases}v &v\leq 2\\ 4-v &v>2\end{cases}$
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  7. #7
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    Re: Simplify

    I am sorry that I did not mention that it was given v <= 2 which for romsek explanation M = v.
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  8. #8
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    Re: Simplify

    Quote Originally Posted by joshuaa View Post
    M = 2 - square_root(u - 3)
    u = (v - 2)^2 + 3

    Prove that M = v.
    I don't think it would be proved..not sure
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