How do I solve Z=the square root of L^2 + R^2 for R.

Printable View

- Dec 6th 2009, 05:38 PMmjfarisSquare Root Problem
How do I solve Z=the square root of L^2 + R^2 for R.

- Dec 6th 2009, 05:44 PMwerepurple
I don't really know what you mean, but I'm guessing you want to make R the subject?

Hence, step by step,

=> Z = (L^2 + R^2)^(1/2)

=> Z^2 = L^2 + R^2

=> R^2 = Z^2 - L^2

=> R = (Z^2 - L^2)^(1/2) <====== - Dec 6th 2009, 08:16 PMNoxide
Wouldn't it be:

=> R =**+/-**(Z^2 - L^2)^(1/2) <======

the +/- because you're solving for all values of R - Dec 6th 2009, 08:19 PMwerepurple
- Dec 6th 2009, 08:23 PMNoxide
actually that sparked a question...

when you are playing with variables in physics and you end up in a situation where x^(n/n) n is even do you put an absolute value symbol around x, or do you just remember that it's the positive case? - Dec 6th 2009, 08:28 PMwerepurple
I don't play with variables in physics, and didn't end up in that situation you desribed. I use that formula to usually find the " change in velocity ", " " change in displacement ", and etc; in other words, I use that formula to solve simple problems most of the time, and it becomes a habit to simply take the positive value as the answer without considering the possible negative value, which is most of the time needless.