Find all real numbers $a$, such that the equation has exactly two solutions:
$$ |x+1|+|2-x|=a^2-1$$
My reasoning:
Since the left side is always positive because of the absolute values, then $a^2-1\gt0$ ? But by solving that condition I get $a\in<-\infty,-1>\cup<1,+\infty>$ but the solution is $a\in<-\infty,-2>\cup<2,+\infty>$; why is this a wrong approach?
How to solve this correctly?
On
Notice that (by the triangle inequality) $|x+1|+|2-x|=a^2-1\geq |x+1+2-x|=3$ so you get $a^2-1\geq 3$ which implies $a^2\geq 4$.
When $-1\leq x\leq 2$ we get that $x+1+2-x=3$ so when $a^2=4$ the equation has infinitely many solutions.
Looking the other cases $x> 2$ and $x< -1$ one gets an unique solution in each interval for each $a^2>4$ (hence for $a^2>4$ there are exactly $2$ solutions).
On
$$|x+1|+|2-x|=a^2-1\\$$ $$ \begin{array}{c||c||c} x<-1 & -1\leq x \leq 2 & x>2\\ \hline \hline |x+1|+|2-x|=a^2-1 & |x+1|+|2-x|=a^2-1 & |x+1|+|2-x|=a^2-1 \\ \hline -x-1+2-x=a^2-1 & x+1+2-x=a^2-1 & x+1+x-2=a^2-1\\ \hline -2x+1 & a^2-1=3 &a^2-1=2x-1\\ \hline a^2-1\geq-2(-1)+1 &a^2=4 & a^2-1 \geq2(2)-1\\ \hline a^2\geq4 & a=\pm2 &a^2\geq 4\\ \hline (-\infty,-2]\cup[2,+\infty) & \pm 2 &(-\infty,-2]\cup[2,+\infty) \end{array} $$ so $$a \in (-\infty,-2]\cup[2,+\infty)$$
On
You noticed correctly that $a^2-1$ must be positive. But keep in mind the other condition that the equation must have exactly two solutions.
Below is the the graph (if you can draw it) of the LHS function:
The RHS is a constant function $a^2-1$, whose graph is a horizontal line.
In order for the two graphs to have two common points (i.e. two solutions): $$a^2-1>3 \Rightarrow a\in(-\infty,-2)\cup(2,+\infty).$$
i would write the equation in the form $$|x+1|+|x-2|=a^2-1$$ 1: for $$x\geq 2$$ we have $$x+1+x-2=a^2-1$$ or $$x=\frac{a^2}{2}\geq 2$$ 2: $$-1\le x<2$$ form here we get $$a^2=4$$ last case: $$x<-1$$ then we get $$-x-1-x+2=a^2-1$$ thus $$x=\frac{2-a^2}{2}$$ and $$\frac{2-a^2}{2}<-1$$ can you finish?