about solving: Absolute value How to solve: $|\sqrt{x-1}-2| + |\sqrt{x-1}-3|=1$.
I would like to know how to solve an absolute value equation when there is a square root sign inside.
 A: Let us call $\sqrt{x-1}$ as $y$. Note that by definition $y \geq 0$. Now we need to find $y$ such that $$\lvert y-2 \rvert + \lvert y - 3 \rvert = 1$$
To solve this lets split into three cases.


*

*$y < 2$. This gives us that $$\lvert y-2 \rvert + \lvert y - 3 \rvert = (2-y) + (3-y) = 5 -2y > 5 -2 \times 2 =1$$ Hence, $y < 2$ is not possible.

*$2 \leq y \leq 3$. This gives us that $$\lvert y-2 \rvert + \lvert y - 3 \rvert = (y-2) + (3-y) = 1$$ Hence, all $y \in [2,3]$ satisfies this.

*$y > 3$. This gives us that $$\lvert y-2 \rvert + \lvert y - 3 \rvert = (y-2) + (y-3) = 2y-5 > 2 \times 3 -5 = 1$$ Hence, $y > 3$ is not possible.


This means that $y \in [2,3]$. Hence, we get that $\sqrt{x-1} \in [2,3]$ i.e. $x - 1 \in [4,9]$. Hence, $$x \in [5,10]$$
A: I recommend the following approach, in this case. Let $y=\sqrt{x-1}$. Now you need only solve the equation $|y-2|+|y-3|=1$--which should have a closed interval's worth of solutions--then given any of the solutions, say $\alpha$, solve the equation $\sqrt{x-1}=\alpha$. In the end, you will obtain a closed interval's worth of solutions.
It is important to note that this approach will not always work! If the equation you'd started with had been $$\left|\sqrt{x+1}-2\right|+\left|\sqrt{x}-3\right|=1,$$ we could not have made the substitution as above.
A: Theorem: Let $X,P,Q \in \mathbb R^n$. Then $X \in \overline{PQ} \iff d(P,X) + d(X,Q) = d(P,Q)$
On the set of real numbers, the distance between numbers $x$ and $y$ is $|x-y|$. So we have this.
Corollary: Let $x,p,q \in \mathbb R$ with $p < q$. Then $p \le x \le q \iff |x-p| + |q-x| = q-p$
In this case, 
\begin{align}
   |\sqrt{x-1}-2| + |\sqrt{x-1}-3|=1 
   &\iff 2 \le \sqrt{x-1} \le 3 \\
  &\iff  4 \le x-1 \le 9 \\
  &\iff 5 \le x \le 10
\end{align}
