The equation √(x + 3-4√(x-1))+ √(x+8-6√(x-1))=1 has

The given is meaningful for $x ≥1$.

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

$⇒\sqrt{(x-1) + 4-4 \sqrt{x-1}} + \sqrt{(x-1)+9-6 \sqrt{x-1}}=1$

$⇒|\sqrt{x-1}-2 |+|\sqrt{x-1}-3|=1$  $[∵\sqrt{x^2}=|x|]$

$⇒|t-2|+|t-3|=1$, where $x-1=t^2$.

as $2≤t≤3$ so $2≤\sqrt{x-1}≤3$

$⇒4≤x-1≤9$

$⇒5≤x≤10$

This equation is satisfied for all values of t lying between 2 and 3. Thus, the given equation is satisfied for all values of x lying between 5 and 10.