How can I solve this absolute value equation? This is the equation:
$|\sqrt{x-1} - 2| + |\sqrt{x-1} - 3| = 1$
Any help would be appreciated. Thanks!
 A: Let $a =\sqrt{x-1}$,
$|a-2|+|a-3|=1$
Check for solutions in the different regions for $a$.
Region 1: $a<2$.
Then we have $(2-a)+(3-a)=1$, i.e. $5-2a=1$, so that $a=2$.
Region 2: $2\leq a\leq 3$. We have $(a-2)+(3-a)=1$. This is true for every such $a$.
Final region 3: $3<a$. We have $(a-2)+(a-3)=1$ so that $a=3$. 
In summary, $2 \leq \sqrt{x-1} \leq 3$.
Thus $5 \leq x \leq 10$.
A: Let $x$ be a solution of the equation. Notice $x \geq 1$, since $\sqrt{x-1}$ has its domain as $x \geq 1$.
If $x \geq 10$, then each of the absolute value is just the term inside (i.e.$|\sqrt{x-1}-2| = \sqrt{x-1}-2$ and similarly $|\sqrt{x-1}-3|  =\sqrt{x-1}-3$) so that the given equation in this case becomes $2\sqrt{x-1}-5=1$. Solve for $x$ you get $x=10$. Thus we have shown that IF there is a solution that is greater than or equal to $10$ then $x$ must be $10$; we have not yet proven that $10$ is a solution. However, we can check that $x=10$ is a solution by plugging it back in. 
If $10 > x \geq 5$ then $|\sqrt{x-1}-2| = \sqrt{x-1}-2$ and $|\sqrt{x-1}-3|  =-\sqrt{x-1}+3$ so that the given equation becomes (after simplification) $1=1$; this does not give us a new information, but we can check that any number $x$ such that $10 > x \geq 5$ satisfies the original equation.
If $5 > x \geq 1$ then $|\sqrt{x-1}-2| = -\sqrt{x-1}+2$ and $|\sqrt{x-1}-3|  = -\sqrt{x-1}+3$, so that the given equation becomes (after simplification) $\sqrt{x-1}=2$, implying $x=5$, contradicting our assumption that $5 > x$. Thus, there cannot be any solution in this case.
Hence, the solution are $5 \leq x \leq 10$.
In general, any time you see an absolute value in a given equation, it's a good idea to divide into cases according to whether each expression in an absolute value is negative or non-negative.
