Question

The equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\text{ has }$

• A. no solution
• B. one solution
• C. two solutions
• D. more than two solutions

Answer 1

Answer: (D)

Put $\sqrt{x-1}=t$
$\Rightarrow x-1=t^2$
or $x=t^2+1$
The given equation reduces to
$\sqrt{t^2+1+3-4t}+\sqrt{t^2+1+8-6t}=1$
where $t\ge 0$.
$\Rightarrow |t-2|+|t-3|=1$
where $t\ge 0$.
This equation will be satisfied if $2\le t\le 3$.
Therefore, $2\le \sqrt{x-1}\le 3$
or $5\le x\le 10$.
$\therefore$ The given equation is satisfied for all values of $x$ lying in $[5,10]$.