about solving: Absolute value

Question

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.

Answer 1

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.

1. $y < 2$. 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. $2 \leq y \leq 3$. 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.
3. $y > 3$. 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, $$x \in [5,10]$$

Answer 2

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.