• Feb 23rd 2011, 08:48 AM
john1875
0 = 400 - 600t^2

transpose for t

I get t = -400/-600^2

• Feb 23rd 2011, 08:54 AM
masters
Quote:

Originally Posted by john1875
0 = 400 - 600t^2

transpose for t

I get t = -400/-600^2

Hi john1875,

Use these steps:

$0=400-600t^2$

$-400=-600t^2$

$\dfrac{-400}{-600}=t^2$

$\dfrac{2}{3}=t^2$

$t=\sqrt{\dfrac{2}{3}}$

Can you rationalize the denominator?

• Feb 23rd 2011, 08:54 AM
Prove It
No, it's not correct. You really should simplify each step before taking on any other step... E.g. $\displaystyle \frac{-400}{-600} = \frac{2}{3}$...

Also, to undo squaring, you don't square both sides. You take the square root of both sides.
• Feb 23rd 2011, 09:30 AM
john1875
Wonder if you could clarify this point? Would I √(2/3) or √2 /3 ?
• Feb 23rd 2011, 09:33 AM
Prove It
It's $\displaystyle \sqrt{\frac{2}{3}}$.