Question

If for $\sqrt{x+3-4\sqrt{(x-1)}}+\sqrt{x+8-6\sqrt{(x-1)}}=1,x\in [n,2n]$, then find $n$ .

Answer 1

Let $\sqrt{(x-1)}=t$
We have $x=t^2+1,t\ge 0$
The given equation reduces in the form
$\sqrt{t^2+4-4t}+\sqrt{t^2+9-6t}=1\Rightarrow |t-2|+|t-3|=1\therefore 2\le t\le 3\Rightarrow 4\le t^2\le 9\Rightarrow 4\le x-1\le 9\Rightarrow 5\le x\le 10$
$\therefore$ Solution of the original equation is $x\in [5,10]$ .