# Thread: help solving for variable

1. ## help solving for variable

$m = \frac{1}{\sqrt{1-k^2}}$

I need to solve for k. Is this the correct steps?

$\sqrt{1-k^2}m = 1$
$\sqrt{1-k^2} = \frac{1}{m}$
$\pm(1-k^2) = \frac{1}{m^2}$

But I get stuck here. How would I isolate for k?

2. Originally Posted by centenial
$m = \frac{1}{\sqrt{1-k^2}}$

I need to solve for k. Is this the correct steps?

$\sqrt{1-k^2}m = 1$
$\sqrt{1-k^2} = \frac{1}{m}$
$\pm(1-k^2) = \frac{1}{m^2}$

But I get stuck here. How would I isolate for k?
i hope the ans below can help u

3. Originally Posted by nikk
i hope the ans below can help u

The one step I don't understand is how you get from

$[\sqrt{1-k^2}]^2 = (\frac{1}{m})^2$

to

$1-k^2 = \frac{1}{m^2}$

I understand that this must be done by the sqrt/power cancelling. But don't i need to add a +- in such a case?

4. Originally Posted by centenial

The one step I don't understand is how you get from

$[\sqrt{1-k^2}]^2 = (\frac{1}{m})^2$

to

$1-k^2 = \frac{1}{m^2}$

I understand that this must be done by the sqrt/power cancelling. But don't i need to add a +- in such a case?

You don't need a plus/minus sign because any real number squared is positive (or 0).

If you were to find m in terms of k you'd need it

5. Originally Posted by centenial

The one step I don't understand is how you get from

$[\sqrt{1-k^2}]^2 = (\frac{1}{m})^2$

to

$1-k^2 = \frac{1}{m^2}$

I understand that this must be done by the sqrt/power cancelling. But don't i need to add a +- in such a case?

+ or - normally the case as below
$
\sqrt 25 = + - 5
$

and not for
$
[\sqrt 25]^2
$

sorry if my explanation,english and latex comman is bad

6. Okay, that makes sense. Thanks so much for your help!

7. Originally Posted by centenial
Okay, that makes sense. Thanks so much for your help!
no problem