# Algebraic formula manipulation - solve for $y$ in the equation $0.25=1-(\sqrt[r]{|y-v|}×s)$

I need to find $y$ in the following formula. I used an online algebraic calculator, but the answer wasn't correct (it omitted the index):

$0.25=1-(\sqrt[r]{|y-v|}×s)$

EDIT

The online calculator gave the following:

$y=v+\dfrac{0.5625}{s^2}$

This doesn't produce correct results, even for r = 2.

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Rearrange to get $$(\textstyle\sqrt[r]{|y-v|}\times s)=1-0.25=0.75.$$ Divide by $s$ to get $$\textstyle\sqrt[r]{|y-v|}=\dfrac{0.75}{s}=\dfrac{3}{4s}.$$ Take the $r$th power to get $$|y-v|=\left(\dfrac{3}{4s}\right)^r.$$ Now, if $y-v$ is non-negative, we have $|y-v|=y-v$, hence $$y=v+\left(\dfrac{3}{4s}\right)^r,$$ and if $y-v$ is negative, we have $|y-v|=v-y$, hence $$y=v-\left(\dfrac{3}{4s}\right)^r.$$ Thus, $$y=v\pm \left(\dfrac{3}{4s}\right)^r$$ and whether the $\pm$ is a $+$ or $-$ depends on the sign of $y-v$. Also, note that this equation doesn't make sense for $s=0$.

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Wow, that was quick. Thank you. –  IanC Jun 27 '12 at 10:08
@IanC: No problem, glad to help. Also, note that it agrees with the online calculator for $r=2$: $$y=v+\left(\frac{3}{4s}\right)^2=v+\frac{3^2}{4^2\times s^2}=v+\frac{9}{16s^2}=v+\frac{0.5625}{s^2}$$ Perhaps you were using values of $v$ where $y-v$ was negative? –  Zev Chonoles Jun 27 '12 at 10:11
I'm not sure what I am doing wrong, because I'm not getting the expected results. I'll figure it out shortly :) –  IanC Jun 27 '12 at 10:13
I checked: $y-v$ is positive. –  IanC Jun 27 '12 at 10:17
All sorted! Thanks again @Zev –  IanC Jun 27 '12 at 10:22