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.

  • $\begingroup$ Is the $x-1$ in the square root or is it just the $x$? Why are $2$ and $3$ in parentheses? $\endgroup$ – A.S Jun 4 '12 at 23:19
  • $\begingroup$ x-1 is in the square root ; 2 and 3 are in parentheses for i just wanted to show that 2,3 are not in the square root $\endgroup$ – mgh Jun 4 '12 at 23:21

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

  • $\begingroup$ how can we say that y is greater than or equal to zero? It can also be like :y<o $\endgroup$ – mgh Jun 5 '12 at 1:00
  • $\begingroup$ @meg_1997: Normally square roots are defined to be positive. If you want both roots, you need an explicit $\pm$ sign. $\endgroup$ – Ross Millikan Jun 5 '12 at 1:11
  • $\begingroup$ @meg_1997 As Ross has already pointed it out, $\sqrt{x-1}$ denotes the positive root of $x-1$. $\endgroup$ – user17762 Jun 5 '12 at 2:18

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.

  • $\begingroup$ Why will it not always work? $\endgroup$ – mgh Jun 5 '12 at 0:58
  • $\begingroup$ The problem lies in the fact that $\sqrt{x+1}$ and $\sqrt{x}$ are not linearly related. If we were to try the substitution $y=\sqrt{x+1}$, then we'd need $\sqrt{x}=\sqrt{y^2-1}$, and that's no good. We face a similar problem if we try the substitution $y=\sqrt{x }$. $\endgroup$ – Cameron Buie Jun 5 '12 at 1:12

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}


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