Quiz Question: Solve for $x$: $9^{2x+1} - 28 (3^{x}) +3 = 0$?

Question

I know this is probably quite basic, but I've got my school mathematics quiz tomorrow and I've honestly drawn a blank with this question:

Find the sum of the roots of the equation:

$9^{2x+1}-28(3^x)+3=0$

Answer in simplest form $a/b$, where $a$ and $b \in \mathbb{Z}$.

Thanks in advance for your help.

Answer 1

The equation doesn't have rational solutions; it would if we fix a probable typo: $$ 9^{2x+1}-28\cdot9^x+3=0 $$ Then, setting $z=9^x$ the equation becomes $9z^2-28z+3=0$, that is, $z=1/9$ or $z=3$.

Thus $x=-1$ or $x=1/2$.

If instead it is $3^{2x+1}-28\cdot3^x+3=0$, setting $z=3^x$ gives again $z=1/9$ or $z=3$. In this case we get $x=-2$ or $x=1$.

Using numeric methods, the given equation has approximate solutions $3^x\approx 1.4222$ or $3^x≈0.10719$, that correspond to $x\approx 0.3206$ or $x\approx-2.0327$.

Answer 2

$0 = 9f = \overbrace{9^{\large 2x+2}}^{\large\rm X^2}\!-28\cdot \overbrace{9^{\Large x+1}}^{\large\rm X}\!+27=\rm (X-\color{#c00}{\bf 1})\,(X-\color{#c00}{27})$

So $\ 9\cdot 9^{\large x}\! = {\rm X} = \color{#c00}{{\bf 1},27}\!\iff\! 9^{\large x} = {\large \frac{1}9},\,3\iff x = -1,\,\large{\frac{1}2} $