Question

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

• A. No solution
• B. One solution
• C. Two solution
• D. More that two solutions

Answer 1

The correct option is D More that two solutions

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

Substitute $\sqrt{x-1}=t\Rightarrow x-1=t^2$ or $x=t^2+1,$ The given equation reduces to

$\sqrt{t^2+1+3-4t}+\sqrt{t^2+1+8-6t}=1$ where, $t\ge 0.$

$\Rightarrow \sqrt{{(t-2)}^2}+\sqrt{{(t-3)}^2}=1$ 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, $2\le \sqrt{x-1}\le 3$ $\Rightarrow 5\le x\le 10.$
$\therefore$ The given equation is satisfied for all values of $x\in [5,10]$

Hence more then two solutions are possible and option (D) is correct.