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Math Help - How to factor this square root out?

  1. #1
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    How to factor this square root out?

    Hi

    So I have this seemingly simple, but I just don't see how this works?

    <br />
\frac{C[(M_1 + M_2) \pm \sqrt{(M_1 + M_2)^2 - (M_1 M_2)(Ka)^2}]}{(M_1 M_2)}<br />

    To

    <br />
\frac{C(M_1 + M_2)}{(M_1 M_2)} \left( 1 \pm \left[ 1 - \frac{(M_1 M_2) (Ka)^2}{2(M_1 + M_2)^2} \right] \right)<br />

    please if you can solve this, explain every step that leads to the result.

    pretty nasty eh? Or maybe I'm just stupid

    Thanks
    Chris
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  2. #2
    Member mybrohshi5's Avatar
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    What are you trying to solve for?

    Or are you trying to get from the first equation to the second?
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  3. #3
    Member mohammadfawaz's Avatar
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    Are you sure that you didn't miss any thing? It is hard to get rid of the square root unless you have some relation between K, a M1, and M2.
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  4. #4
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    Hello, chutsu!

    There are two typos in your problem.
    I don't see where that "2" comes from.
    Also, you left out the square-root in the second expression.


    How we get from: . C\cdot \frac{(M_1 + M_2) \pm \sqrt{(M_1 + M_2)^2 - M_1 M_2(Ka)^2}}{M_1 M_2}

    . . . to: . C\cdot\frac{(M_1 + M_2)}{M_1 M_2}\cdot \left[ 1 \pm \sqrt{ 1 - \frac{M_1M_2(Ka)^2}{(M_1 + M_2)^2}}\, \right]

    Under the radical, we have: . (M_1+M_2)^2 - M_1M_2(Ka)^2


    Multiply the second term by \frac{(M_1+M_2)^2}{(M_1+M_2)^2}\!:

    . . (M_1+M_2)^2 \;-\; \frac{M_1M_2(Ka)^2}{1}\cdot\frac{(M_1+M_2)^2}{(M_1  +M_2)^2} \;\;= .  (M_1+M_2)^2 \;-\; \frac{M_1M_2(Ka)^2\cdot(M_1+M_2)^2}{(M_1+M_2)^2}

    Factor: . (M_1+M_2)^2\cdot\left[1 - \frac{M_1M_2(Ka)^2}{(M_1+M_2)^2}\right]

    Take the square root: . \sqrt{(M_1+M_2)^2\cdot\left[1 - \frac{M_1M_2(Ka)^2}{(M_1+M_2)^2}\right]} . =\;(M_1+M_2)\cdot\sqrt{1 - \frac{M_1M_2(Ka)^2}{(M_1+M_2)^2}}



    Substitute into the original expression:

    . . C\cdot\frac{(M_1+M_2) \pm\left[(M_1+M_2)\cdot\sqrt{1 - \dfrac{M_1M_2(Ka)^2}{(M_1+M_2)^2}}\right]} {M_1M_2}


    Factor out \frac{M_1+M_2}{M_1M_2}\!:\quad C\cdot\frac{M_1+M_2}{M_1M_2}\cdot   \left[1 \pm\sqrt{1 - \frac{M_1M_2(Ka)^2}{(M_1+M_2)^2}} \right]

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  5. #5
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    I think I got it, the square root disappears because I can do a Binomial Approximation which states:

    <br />
(1-x)^n = 1 - nx<br />

    Therefore solving the two missing "2"s you were saying. Am i correct?

    P.S. This was part of a physics problem I have (Solid State Physics), which the algebra was kind of horrible to solve as you can see from the above...
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  6. #6
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    Quote Originally Posted by chutsu View Post
    I think I got it, the square root disappears because I can do a Binomial Approximation which states:

    <br />
(1-x)^n = 1 - nx<br />

    Therefore solving the two missing "2"s you were saying. Am i correct?

    P.S. This was part of a physics problem I have (Solid State Physics), which the algebra was kind of horrible to solve as you can see from the above...
    Yes,

    \sqrt{\left(1-\left[\frac{1}{2}\right]\frac{M_1M_2(K_a)^2}{(M_1+M_2)^2}\right)^2}=

    \sqrt{\left(1-\frac{1}{2}\frac{M_1M_2(K_a)^2}{(M_1+M_2)^2}\right  )\left(1-\frac{1}{2}\frac{M_1M_2(K_a)^2}{(M_1+M_2)^2}\right  )}

    =\sqrt{1-\frac{M_1M_2(K_a)^2}{(M_1+M_2)^2}+\frac{1}{4}\ \frac{M_1^2M_2^2(K_a)^4}{(M_1+M_2)^4}}

    and you may discard the final term if it is negligible.
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