Find Real Roots of $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$

[kaliprasad]

find the real roots of the equation

$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1 $

 

[MarkFL]

My solution:

Spoiler

Let $u=\sqrt{x-1}$ and the equation becomes:

$$|u-2|+|u-3|=1$$

For which we find:

$$2\le u\le3$$

or:

$$5\le x\le10$$

 

[soroban]

Hello, kaliprasad!

Find the real roots: $\:\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}\:=\:1$

Spoiler

We note that:
$\quad \begin{array}{ccccc}x + 3 - 4\sqrt{x-1} &=& (x-1) - 4\sqrt{x-1} + 4 &=& (\sqrt{x-1}-2)^2 \\ x+8 - 6\sqrt{x-1} &=& (x-1) - 6\sqrt{x-1} + 9 &=& (\sqrt{x-1} - 3)^2 \end{array}$

The equaton becomes: $\:\sqrt{(\sqrt{x-1}-2)^2} + \sqrt{(\sqrt{x-1}- 3)^2} \;=\;1$

$\quad \sqrt{x-1} - 2 + \sqrt{x-1}-3 \;=\;1 \quad\Rightarrow\quad 2\sqrt{x-1} \:=\:6 $

$\quad\sqrt{x-1} \:=\:3 \quad\Rightarrow\quad x-1 \:=\:9 \quad\Rightarrow\quad x \:=\:10$