Find all values of $m$ such that the equation $3^{x^2 + 2mx + 4m - 3} - 2 = \left|\dfrac{m - 2}{x + m}\right|$ has two distinct roots on $[-4; 0]$.

Question

Consider the equation $3^{x^2 + 2mx + 4m - 3} - 2 = \left|\dfrac{m - 2}{x + m}\right|$. All values of $m$ such that the above equation has two distinct roots on $[-4; 0]$ are $$\begin{aligned} &&A. \, m \in [1; 3] &&B. \, m \in (1; 3)\\ &&C. \, m \in [1; 3] \setminus \{2\} &&D. \, m \in (-\infty; 1] \cup [3; +\infty) \end{aligned}$$

[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)

By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]

Let's do this question by process of elimination first. For $m = 2$, the equation becomes $$3^{x^2 + 4x + 5} - 2 = 0 \implies x^2 + 4x + (5 - \log_32) = 0$$

And since the simplified discriminant (that's what it's called here) is $\Delta = 2^2 - (5 - \log_32) = \log_32 - 1 < 0$, the above equation doesn't have any roots for $m = 2$, which means we can eliminate choices $A$ and $B$.

Next up, for $m = 0$, the equation becomes $$\begin{aligned} 3^{x^2 - 3} - 2 = \left|\dfrac{-2}{x}\right| &\iff \dfrac{3^{x^2}}{27} = 2 \times \left(1 + \dfrac{1}{|x|}\right) \iff \dfrac{3^{x^2}}{54} = \dfrac{|x| + 1}{|x|}\\ &\iff 3^{x^2}|x| = 54(|x| + 1) \iff (3^{x^2} - 54)|x| - 54 = 0 \end{aligned}$$

Consider function $f(x) = (3^{x^2} - 54)|x| - 54, x \in [-4; 0]$. Using the TABLE function of my calculator, (isn't technology amazing?) we have this table below, (who could have guessed?)

It seems that there's only one root on $[-4; 0]$, which is $x = -2$. Therefore, $m = 0$ is not one of the viable values.

It can be concluded that the correct answer is $C. m \in [1; 3] \setminus \{2\}$.

"You already found the correct answer. What're you asking then?" You might think. Well, of course, I want to know how to actually solve this problem, still within the time limit required.

Do I actually know what to do first? Uhhh, hmmm~ I wish...

Anyhow, thanks for reading, (and even more so if you could help), have a great tomorrow, everyone~

Answer 1

As mentioned in a comment by eyeballfrog, the substitutions $n = m - 2$ and $y = x + m = x + n + 2$ will simplify the equation to $$ 3^{y^2 - n^2 + 1} - 2 = \Bigl\lvert \frac{n}{y} \Bigr\rvert. \tag{*} \label{eq} $$

We notice that, when $y = n \neq 0$, $\eqref{eq}$ becomes $$ 3^{1} - 2 = 1, $$ so $x = -2$ is a root whenever $m \neq 2$.

Similarly, $\eqref{eq}$ holds when $y = -n \neq 0$, so $x = -2n - 2 = -2m + 2$ is also a root when $m \neq 2$.

Notice that the LHS of $\eqref{eq}$ is strictly increasing and the RHS of $\eqref{eq}$ is strictly decreasing on $y \in (0, +\infty)$, so it has at most one root in this range. Also, both sides of $\eqref{eq}$ are invariant under the transformation $y \mapsto -y$, so its root in $(-\infty, 0)$ is exactly the root in $(0, +\infty)$ but negated.

In conclusion, the equation has

• no roots if $m = 2$ (as you have shown);
• the two (distinct) roots $x = -2$ and $x = -2m + 2$ otherwise.

The rest should be straightforward.

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Here's an animation on GeoGebra.