help solving for variable

**centenial** · March 26th 2010, 11:15 AM

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

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?

**nikk** · March 26th 2010, 11:31 AM

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

**centenial** · March 26th 2010, 11:43 AM

Originally Posted by nikk

i hope the ans below can help u

Thank you for your quick and helpful reply.

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?

Thanks for your explanation!

**e^(i*pi)** · March 26th 2010, 11:50 AM

Originally Posted by centenial

Thank you for your quick and helpful reply.

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?

Thanks for your explanation!

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

**nikk** · March 26th 2010, 11:53 AM

Originally Posted by centenial

Thank you for your quick and helpful reply.

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?

Thanks for your explanation!

+ or - normally the case as below
$<br /> \sqrt 25 = + - 5<br />$

and not for
$<br /> [\sqrt 25]^2 <br />$

sorry if my explanation,english and latex comman is bad

**centenial** · March 26th 2010, 11:54 AM

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

**nikk** · March 26th 2010, 11:58 AM

Originally Posted by centenial

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

no problem

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