What I would like to show is that indeed we have only one real solution for $|x+1|=|3^x+5|$, call it $x_0$, and that the solution to the inequality is $(x_0,+\infty)$. Also, $x_0 \approx -6.001$.
Your idea to get rid of absolute values is a good one and your second approach works quite nicely.
You correctly conclude that since $3^x + 5 > 0$, you can just remove absolute value on the RHS of inequality.
For $|x+1|$ it looks like you have correct idea to look at cases, but your notation is incorrect. The correct cases to consider are:
- $x+1\geq 0$,
- $x+1<0$.
In the 1st case, the inequality becomes $x+1<3^x + 5$. I claim that this is true for all $x\geq -1$. This follows from well known inequality $$e^x \geq x + 1,\quad \forall x\in\mathbb R.$$
If you never saw this, just plot the graphs of $e^x$ and $x+1$. You will see that the line $y = x + 1$ is tangent to the graph of $e^x$ at $x=0$. Let us use it on your problem: $$3^x = e^{x\ln 3} \geq x\ln 3 + 1 \implies 3^x + 5 \geq x\ln 3 + 6,$$
and you can easily see that $x\ln 3 + 6 > x + 1$ for $x\geq - 1$, which proves that our inequality is true on $[-1,\infty)$.
The 2nd case is more complicated. Now the inequality becomes
$$-x-1 < 3^x + 5 \iff 3^x + x + 6 > 0.$$ First let us observe that the function $3^x + x + 6$ is strictly increasing. That means that the equation $3^x + x + 6 = 0$ either has no solutions or if it has solutions, it has only one solution. If we plug in $x = -6$, we get $3^{-6}$ which is positive, and if we plug in $x = -7$ we get $3^{-7} - 1$ which is negative. That means that (by continuity) that there exists $x_0 \in (-7,-6)$ such that $3^{x_0} + x_0 + 6 = 0$. Since $x_0< -1$, it is also the unique solution to $|x+1| = |3^x + 5|$ as I claimed at the start. Anyway, we now know that $$3^x + x + 6 > 0 \iff x\in (x_0,+\infty),$$ and taking into account that we are in our 2nd case, $$|x+1| < |3^x + 5|,\quad\forall x\in (x_0,-1).$$
Taking the union of our two cases, we conclude that the solution to our inequality is $(x_0,+\infty)$.
Hopefully, the above is not too hard to follow. What remains is to approximate $x_0$. As we already saw above, $3^x + x + 6$ evaluates to $3^{-6}$ for $x = -6$, which is quite close to $0$, so we expect $x_0$ to be close to $-6$ and more precisely, $x_0$ is a bit less than $-6$.
To get better approximations, we could use numerical methods, but since this is pre-calculus, I don't want to get into that.
I will get into something that is called Lambert W function, which you could argue is even less appropriate, but I think the arithmetic that we will perform with it is not too advanced.
To explain what Lambert W function does we need to observe equation of the form $$xe^x = a.$$ This equation might have $0$, $1$ or $2$ real solutions. If $a\geq 0$ then the equation $xe^x = a$ has unique nonnegative solution, and we will denote it by $x = W(a)$. Compare this to the equation $x^2 = a$ and how we denote one of its solutions with $x = \sqrt a$. We don't really know precise value of $W(a)$ in most cases, but then again, we don't know the precise value of $\sqrt a$ in most cases either. Luckily, we know how to approximate both $\sqrt a$ and $W(a)$.
But, let us return to our equation $3^x + x + 6 = 0.$ I will manipulate it so that we may use Lambert W:
\begin{align}
3^x + x + 6 = 0 & \iff -(x+6) = e^{x\ln 3}\ /\cdot e^{-(x+6)\ln 3}\\
& \iff -(x+6)e^{-(x+6)\ln 3} = e^{x\ln 3-(x+6)\ln 3}\\
& \iff -(x+6)e^{-(x+6)\ln 3} = 3^{-6}\ /\cdot \ln 3\\
& \iff -(x+6)\ln3 \cdot e^{-(x+6)\ln 3} = 3^{-6}\ln 3.
\end{align}
Now, if we substitute $y = -(x+6)\ln 3$, the last line becomes $ye^y = 3^{-6}\ln 3$ and we conclude that $y = W(3^{-6}\ln 3)$ (note that $3^{-6}\ln 3>0$). Substituting $x$ back, we can now easily calculate that $$x = -6 - \frac 1{\ln 3} W(3^{-6}\ln 3),$$
which is our $x_0$ from before.
Finally, it happens to be that $W(a) \approx a$ when $a$ is close to $0$. This is analogous to $\sin a \approx a$ when $a$ is close to $0$, if you ever saw that one, without going into reasons why that might be true. Now, since $3^{-6}\ln 3 \approx 0.00150701$, we have $W(3^{-6}\ln 3) \approx 3^{-6}\ln 3$. With this we can now approximate $$x_0 = -6 - \frac 1{\ln 3} W(3^{-6}\ln 3) \approx - 6 - \frac{3^{-6}\ln 3}{\ln 3} = -6 - 3^{-6} \approx -6.0013717.$$
This approximation is quite good. You can check that Wolfram Alpha approximates $x_0$ to be $-6.00137$.
I assume you are asking how to solve inequalities by squaring both sides in general regardless of this particular example.
You can consider this as a general rule:
You can never square both sides of an inequality, unless both sides are positive.
You already noticed why this rule must be followed, when you tried to square both sides of this inequality:
$$2 > -3$$
So when you're asking: "When to flip the inequality sign?". The answer is:
You never flip the inequality sign, when squaring both sides in accordance with the aforementioned rule.
Then you may ask: "How do I solve an inequality which contains a square root (needs squaring in order to solve it), and which does not have 2 positive sides?"
There are 3 elementary cases for an inequality with a square root:
- Both sides are positive:
Square both sides of the inequality, and continue accordingly, ex:
\begin{align*}
5 &> \sqrt{-x} &&\text{:Defined only when } x\leq0\\
25 &> -x &&\text{:Square both sides}\\
x &> -25 &&\text{:The Pre-Solution}\\
-25 &< x \leq 0 &&\text{:Final solution is the intersection of the definition and the pre-solution}\\
\end{align*}
- Both sides are negative:
Multiply both sides by $-1$ (and flip the inequality sign), and it's reduced to case 1. Ex:
\begin{align*}
-5 &< -4\sqrt{1+x^2} &&\text{:Defined when } x \in \mathbb{R}\\
5 &> 4\sqrt{1+x^2} &&\text{:Multiply both sides by }-1\\
25 &> 16(1+x^2) &&\text{:Square both sides}\\
9 &> 16x^2\\
x^2 &< \frac{9}{16}\\
-\frac{3}{4} < x &< \frac{3}{4} &&\text{:Pre-solution}\\
x \in &\left(-\frac{3}{4},\frac{3}{4}\right) &&\text{:Final solution}
\end{align*}
- One side is positive and one is negative:
Here the inequality is either impossible or always true, because positive is always greater than negative, ex:
\begin{align*}
-3&<\sqrt{x}&&\text{:Defined when }x\geq0 \text{ and always true when }x\geq0\\
5&<-\sqrt{1-x^2}&&\text{:Impossible}
\end{align*}
P.S, When you have a side that is not entirely positive or negative, you divide the domain of definition into parts where it is either entirely positive or entirely negative, to reduce the problem to one or more of the cases above. Ex:
$$
x > \sqrt{x+6} \quad \text{:Defined when }x \geq -6\\
$$
Here, LHS is negative when $x \in [-6,0)$, and positive when $x \in [0,+\infty)$.
So, when $x \in [-6,0)$ we have case 3, and the inequality is impossible.
When $x \in [0,+\infty)$ both sides are positive and we have case 1, so we can square both sides:
\begin{align*}
x^2 &> x+6 && \text{:Square both sides}\\
x^2-x-6 &> 0\\
\end{align*}
\begin{align*}
&x \in (-\infty,-2) \cup (3,+\infty) &&\text{:Pre-solution}\\
&x \in (3,+\infty) &&\text{:Final solution is the intersection of the Pre-solution with }[0,+\infty)
\end{align*}
Now, to address your example:
$$x>-2\sqrt{2x-4} \qquad \text{:Defined when }x \geq 2$$
Here, when $x \geq 2$ LHS is positive, and RHS is negative. So this is case 3 (always true when defined).
So this inequality is always true when $x \geq 2$. (No need to square both sides)
$$x \in [2,+\infty) \quad \text{:Final Solution}$$
Best Answer
Denote:
$x^4-3x^2+5 = a - 7/2$
Then:
$x^4-3x^2+12 = a + 7/2$
Now you want to solve this one:
$\sqrt{a-7/2} + \sqrt{a + 7/2} = 7$
This is quite symmetrical and nice to work with.
Raise it to power $2$ and proceed, should be trivial from there.
At the end do a direct check to see if the values you found for $a$ give rise to valid roots for $x$. For example, if you get a solution for $a$ smaller than $7/2$, that obviously does not work i.e. does not give you any solutions for $x$.