# Rearrange Formula Past Exam Question

• Apr 13th 2011, 02:20 AM
dementedsquirrel
Rearrange Formula Past Exam Question
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. (Doh)

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? (Worried)

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. (Speechless)
• Apr 13th 2011, 02:27 AM
technopersia

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

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

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

continueing:

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

$\displaystyle t=(\sqrt((\frac{8}{F})^2 + 6))$
• Apr 13th 2011, 02:30 AM
dementedsquirrel
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.
• Apr 13th 2011, 02:53 AM
Carlow52
This has been superceded
• Apr 13th 2011, 02:59 AM
dementedsquirrel
Quote:

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?
• Apr 13th 2011, 03:20 AM
technopersia

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

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

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

continueing:

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

$\displaystyle t=(\sqrt((\frac{8}{F})^2 + 6))$
• Apr 13th 2011, 03:28 AM
Carlow52
Again superceded
• Apr 13th 2011, 03:49 AM
dementedsquirrel
Perfect! (Nod)
Thank you very much.
I can't believe I couldn't get it in the first place (Sadsmile)
• Apr 13th 2011, 03:55 AM
technopersia
• Apr 13th 2011, 04:08 AM
dementedsquirrel
No problem.
• Apr 13th 2011, 04:14 AM
e^(i*pi)
$\displaystyle t= \pm \sqrt\left({\dfrac{8}{F}\right)^2 + 6}$. Alternatively $\displaystyle |t|= \sqrt\left({\dfrac{8}{F}\right)^2 + 6}$
To fit the original domain then $\displaystyle t - 6 > 0$ and from your rearranging $\displaystyle F \neq 0$