# 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!

## 2 Answers

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: $$3. We have $$(a-2)+(a-3)=1$$ so that $$a=3$$.

In summary, $$2 \leq \sqrt{x-1} \leq 3$$.

Thus $$5 \leq x \leq 10$$.

• I think you have the equation wrong. There should be a one on the right side not a three. – Shervin Sorouri Apr 5 at 6:57
• yeah it's being sorted – George Dewhirst Apr 5 at 6:57
• For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$. – Dbchatto67 Apr 5 at 6:57

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.