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.
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.
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.
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, $x \in [5,10]$
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.
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}