Week #57 Problem: Solving |√(x-1)-2| + |√(x-1)-3| = 1

[Jameson]

Solve $$\lvert \sqrt{x-1}-2 \rvert + \lvert \sqrt{x-1} - 3 \rvert = 1$$
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[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone
3) kaliprasad
4) soroban
5) Sudharaka
6) Reckoner

Solution (from Reckoner):

Spoiler

The expression inside the absolute value bars on the left is negative when $1 \leq x < 5$ and nonnegative for $x\geq5.$ Similarly, the expression inside the second set of absolute value bars is negative when $1 \leq x < 10$ and nonnegative for $x\geq10.$ This gives us three cases to consider.

If $1\leq x\leq 5,$ we have
\begin{align*}
\left|\sqrt{x - 1} - 2\right| + \left|\sqrt{x - 1} - 3\right| = 1
&\Leftrightarrow -\left(\sqrt{x - 1} - 2\right) - \left(\sqrt{x - 1} - 3\right) = 1\\
&\Leftrightarrow -2\sqrt{x-1} + 5 = 1\\
&\Leftrightarrow \sqrt{x - 1} = 2\\
&\Leftrightarrow x = 5,
\end{align*}
and so $5$ is the only solution in this first interval.

If $5\leq x\leq10,$ we have
\begin{align*}
\left|\sqrt{x - 1} - 2\right| + \left|\sqrt{x - 1} - 3\right| = 1
&\Leftrightarrow \left(\sqrt{x - 1} - 2\right) - \left(\sqrt{x - 1} - 3\right) = 1\\
&\Leftrightarrow 1 = 1
\end{align*}
so the equation is true for all $x$ in this interval.

Finally, if $x\geq10,$ we have
\begin{align*}
\left|\sqrt{x - 1} - 2\right| + \left|\sqrt{x - 1} - 3\right| = 1
&\Leftrightarrow \left(\sqrt{x - 1} - 2\right) + \left(\sqrt{x - 1} - 3\right) = 1\\
&\Leftrightarrow 2\sqrt{x-1} - 5 = 1\\
&\Leftrightarrow \sqrt{x - 1} = 3\\
&\Leftrightarrow x = 10.
\end{align*}

Therefore, the original equation is satisfied for all $x$ in the interval $[5, 10],$ and there are no solutions outside this interval.

Alternate solution (from soroban):

Spoiler

$$\text{Solve: }\:|\sqrt{x-1}-2| + |\sqrt{x-1}-3| \:=\:1$$Let $$u \,=\,\sqrt{x-1}$$

We have: .[/color]$$|u - 2| + | u -3| \:=\:1$$

We want a number $$u$$ whose sum of distances
. . [/color]from 2 and 3 is equal to 1.

We find that $$u$$ lies on this interval:
. . [/color]$$\begin{array}{ccccc} --- & \bullet & === & \bullet & --- \\ & 2 && 3 \end{array}$$

That is: .[/color]$$2 \;\le u\;\le 3$$

. . . [/color]$$2 \;\le\; \sqrt{x-1}\;\le\;3$$

. . . . [/color]$$4 \;\le \; x-1 \;\le\;9$$

. . . . . [/color]$$5 \;\le\;x\;\le\;10$$