Rearrange Formula Past Exam Question

**dementedsquirrel** · April 13th 2011, 02:20 AM

I'm really sorry for this because I know it's not the hardest of formulas, but I have a complete blank on how to do it.

I need to make t the subject of the formula.

F= 8 / root(t^2-6)

^2 being squared.

Don't worry, it's not for any homework or assignment, I'm revising for my exam in 3 weeks time. I know it's silly that I can't do this at such a late stage, but can anyone help?

I thought I had to times both sides by root(t^2-6) and divide both sides by F first, but I could easily be wrong.

**technopersia** · April 13th 2011, 02:27 AM

If this is your formula:

$F=\frac{8}{\sqrt{t^2-6}}$

then multiplying both sides by $\sqrt{t^2-6}$ and deviding both sides by F gives you:

$\sqrt{t^2-6}=\frac{8}{F}$

continueing:

$t^2=(\frac{8}{F})^2 + 6$

$t=(\sqrt((\frac{8}{F})^2 + 6))$

**dementedsquirrel** · April 13th 2011, 02:30 AM

Yeah I thought I should do that, but I didn't know what to do after to get t on it's own. But thanks

EDIT.
Sorry, the bottom bits didn't show up when I first read your post. Ignore this one.

**Carlow52** · April 13th 2011, 02:53 AM

This has been superceded

**dementedsquirrel** · April 13th 2011, 02:59 AM

Originally Posted by Carlow52

Is the move from first line to 2nd line correct here?

Just wondering as it looks like the LHS has been squared and the RHS 'rooted'

That is a point. Are you not supposed to square it? Or am I just talking baloney?

**technopersia** · April 13th 2011, 03:20 AM

You are right, my bad!

If this is your formula:

$F=\frac{8}{\sqrt{t^2-6}}$

then multiplying both sides by $\sqrt{t^2-6}$ and deviding both sides by F gives you:

$\sqrt{t^2-6}=\frac{8}{F}$

continueing:

$t^2=(\frac{8}{F})^2 + 6$

$t=(\sqrt((\frac{8}{F})^2 + 6))$

**Carlow52** · April 13th 2011, 03:28 AM

Again superceded

**dementedsquirrel** · April 13th 2011, 03:49 AM

Perfect!

Thank you very much.
I can't believe I couldn't get it in the first place

**technopersia** · April 13th 2011, 03:55 AM

sorry about the first mistake.

**dementedsquirrel** · April 13th 2011, 04:08 AM

No problem.

**e^(i*pi)** · April 13th 2011, 04:14 AM

Don't forget your negative root!

$t= \pm \sqrt\left({\dfrac{8}{F}\right)^2 + 6}$ . Alternatively $|t|= \sqrt\left({\dfrac{8}{F}\right)^2 + 6}$

To fit the original domain then $t - 6 > 0$ and from your rearranging $F \neq 0$

**dementedsquirrel** · April 13th 2011, 04:16 AM

Thank you.

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