Question

3x=sqr of 9x+2 over 2
been stuck on this one for a while

Answer 1

\[3x=\sqrt{(9x+2)/2}\]
\[9x^2=(9x+2)/2 \]
\[18x^2=9x+2 \]
\[18x^2−9x−2=0 \]
\[(3x−2)(6x+1)=0 \]
\[x=2/3,−1/6\]

However, remember that we started with a radical, and a radical can only produce a positive number. If we use x=-1/6, we see the radical must equal -1/2 (=3x), which is not possible. Therefore, the only real solution is 2/3. The reason why there's a second solution is because as part of solving the equation, I squared both sides, introducing a second possible solution.

Answer 2

:) oh thank you!