I believe there was an error with the text of the following question, but I just want to make sure my math is perfect here and the text is really wrong:

- Be a a real number so that \(\displaystyle 3y^2 - y + a = 0\) have double roots. So, the solution to the following equation \(\displaystyle 3^{2x+1} - 3^x + a = 0\) is...?

Well, of course the second equation is exactly the same as first, with \(\displaystyle y=3^x\)

Solving, we got: \(\displaystyle 3^x= \frac{1+-sqrt{1-12a}}{6}\)

As none of the answer alternatives is given in terms of a, I suspect the text of the question was wrong and actually that \(\displaystyle 1-12a=0\), giving thus the answer \(\displaystyle x=-\log_{3}{6}\), which is the answer on the answer sheet.

So the text of the question is really:

- Be a a real number so that \(\displaystyle 3y^2 - y + a = 0\)

**DOESN'T**have double roots. So, the solution to the following equation \(\displaystyle 3^{2x+1} - 3^x + a = 0\) is...?

Thanks