# Math Help - How to factor this square root out?

1. ## 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

2. What are you trying to solve for?

Or are you trying to get from the first equation to the second?

3. 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.

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

5. 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...

6. 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.