Could someone check my soln to -9(x-3)^2 = -7

Could someone just check to see if I did this right? I would appreciate it.
Solve:

-9(x-3)^2 = -7

x-3 = sqrt7/9

x= 3 +/- sqrt63 / 9

x= 27 +/- sqrt63 over 9

\(\displaystyle \L -9(x-3)^2 = -7\)

\(\displaystyle \L (x-3)^2 = \frac{7}{9}\)

\(\displaystyle \L x = 3\)\(\displaystyle \tiny +/-\)\(\displaystyle \L \sqrt{\frac{7}{9}}\)

k, guess i was wrong, lol. Quick question though... how do i know when I am supposed to rationalize the denominator? That's what I did with this one. I thought teachers didnt like it when there was a sqrt in the denominator? Furthermore, wouldnt the denominator be 3 because the of sqrt9?

I am confused

When you get to later math courses, you will discover that the instructors no longer care if there are radicals in the denominator. The tutor probably just forgot that, at your stage, your instructor might "have a hissy fit" about the sqrt[9].

In all cases, however, yes, the simplification from sqrt[9] to 3 should be done. (The tutor was leaving a little bit for you to do. Good tutors don't give the whole answer!) :wink:

Eliz.

yes... \(\displaystyle \L \sqrt{\frac{7}{9}} = \frac{ \sqrt{7}}{3}\)but then the 3 must be outside the radical..

\(\displaystyle \L let \sqrt{\frac{7}{9}}\) be two radicals.

\(\displaystyle \L \frac{\sqrt{7}}{\sqrt{9}} = \frac{\sqrt{7}}{3}\)