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.

$9(3^{4x})-28(3^x)+3=0$ = $9a^4-28a +3=0$, I'm not sure where to go from here? We haven't done solving powers above 3 yet :( — Peter O'Flynn, Jan 22 '19 at 22:29
Hey there! Out of curiosity, what grade are you in? Trying to figure out how difficult this question is — J. Dionisio, Jan 22 '19 at 22:44
Well i'm in school in Ireland so we don't have a grade system - but i'm just 18 if that helps? I'll be going to college (hopefully) next year. — Peter O'Flynn, Jan 22 '19 at 22:45
Wolframalpha shows two solutions, approximately $3^x\approx 1.4222$ and $3^x\approx 0.10719$; no way you can express the sum of the roots as a rational number. — egreg, Jan 22 '19 at 22:49
I strongly suspect there's an error in the question, and that it was meant to be $9^{x+1}$ or $3^{2x+1}$, so that the transformed equation would be quadratic. Solving... either works. — jmerry, Jan 22 '19 at 22:53
Possibly, the answer they give is $-1/2$ and its the exact powerpoint they used at the competition so I thought it was unlikely. Sorry about that guys. — Peter O'Flynn, Jan 22 '19 at 22:58

score 2 · Accepted Answer · answered Jan 22 '19 at 22:56

2

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 $x\approx-2.0327$.

answered Jan 22 '19 at 22:56

egreg

238,574

Yep this is correct, the answer they give is $-1/2$. Thanks for that, I would have been up all night otherwise. – Peter O'Flynn Jan 22 '19 at 23:02
@PeterO'Flynn, it would be nice to vote for the answer, specially if it was useful. – NoChance Jan 22 '19 at 23:09
@NoChance, I did, but it says that since I have under 15 reputation it doesn't display it publicly. – Peter O'Flynn Jan 23 '19 at 18:05

Bill Dubuque · Answer 2 · 2019-01-22T23:28:21.267

1

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

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} $

edited Jan 22 '19 at 23:28

answered Jan 22 '19 at 23:18

Bill Dubuque

272,048

1

Presuming $\ 3^{\large x}\ $ is a typo for $,9^{\large x},,$ as seems to be true based on your comments on egreg's answer. – Bill Dubuque Jan 22 '19 at 23:19
The initial scaling by $9$ is the AC method, to make the polynomial monic (lead coef $=1)\ \ $ – Bill Dubuque Jan 22 '19 at 23:35
@NoChance What do you mean by "type is fixed"? The Roman typeface is used by choice to help avoid confusion between $,x,$ and $,{\rm X} = 9^{\large x+1}.\ $ – Bill Dubuque Jan 24 '19 at 14:06
+1 indeed, Thank you for your reply you are correct, I confused this with something else. – NoChance Jan 24 '19 at 14:22

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

2 Answers2