Question

The number of integral values of $x$ for which the expression
$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ holds true is

Answer 1

$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$
$\Rightarrow \sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1$
$\Rightarrow \sqrt{|\sqrt{(x-1)}-2{|}^2}+\sqrt{|\sqrt{(x-1)}-3{|}^2}=1$
$\Rightarrow |\sqrt{(x-1)}-2|+|\sqrt{(x-1)}-3|=1$
$\Rightarrow |\sqrt{(x-1)}-2|+|\sqrt{(x-1)}-3|=(\sqrt{(x-1)}-2)-(\sqrt{(x-1)}-3)$
$\Rightarrow \sqrt{(x-1)}-2\ge 0$ and $\sqrt{(x-1)}-3\le 0$
$\Rightarrow 2\le \sqrt{x-1}\le 3$
$\Rightarrow 4\le x-1\le 9$
$\Rightarrow 5\le x\le 10$