# How to factor this square root out?

• May 11th 2010, 12:57 PM
chutsu
How to factor this square root out?
Hi

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

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

To

$
\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
• May 11th 2010, 01:47 PM
mybrohshi5
What are you trying to solve for?

Or are you trying to get from the first equation to the second?
• May 11th 2010, 02:18 PM
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.
• May 11th 2010, 03:07 PM
Soroban
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.

Quote:

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

• May 11th 2010, 03:30 PM
chutsu
I think I got it, the square root disappears because I can do a Binomial Approximation which states:

$
(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...
• May 12th 2010, 04:19 AM
Quote:

Originally Posted by chutsu
I think I got it, the square root disappears because I can do a Binomial Approximation which states:

$
(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,

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

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

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

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