How to factor this square root out?

May 2010
13
0
Hi

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

\(\displaystyle
\frac{C[(M_1 + M_2) \pm \sqrt{(M_1 + M_2)^2 - (M_1 M_2)(Ka)^2}]}{(M_1 M_2)}
\)

To

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

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
 
Sep 2009
230
7
What are you trying to solve for?

Or are you trying to get from the first equation to the second?
 
Feb 2010
100
27
Lebanon - Beirut
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.
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
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: . \(\displaystyle C\cdot \frac{(M_1 + M_2) \pm \sqrt{(M_1 + M_2)^2 - M_1 M_2(Ka)^2}}{M_1 M_2}\)

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

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


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

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

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

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



Substitute into the original expression:

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


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

 
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May 2010
13
0
I think I got it, the square root disappears because I can do a Binomial Approximation which states:

\(\displaystyle
(1-x)^n = 1 - nx
\)

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...
 
Dec 2009
3,120
1,342
I think I got it, the square root disappears because I can do a Binomial Approximation which states:

\(\displaystyle
(1-x)^n = 1 - nx
\)

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,

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

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

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

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