Problem of the week #57 - April 29th, 2013

[Jameson]

Solve \(\displaystyle \lvert \sqrt{x-1}-2 \rvert + \lvert \sqrt{x-1} - 3 \rvert = 1\)
--------------------
Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[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

[tex]\text{Solve: }\:|\sqrt{x-1}-2| + |\sqrt{x-1}-3| \:=\:1 [/tex]


Let [tex]u \,=\,\sqrt{x-1}[/tex]

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

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

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

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

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

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

. . . . . [tex]5 \;\le\;x\;\le\;10[/tex]