QJEEM19Ma463a: JEE Main 2019 Mathematics - Class 11 JEEM19Ma463a Solution: The sum of the solutions of the equation (x > 0) is equal to:

Case 1 : When $\sqrt x \ge 2$

then $\left| {\sqrt x - 2} \right| = \sqrt x - 2$

$\therefore$ The given equation becomes,

$\left( {\sqrt x - 2} \right)$ + $\sqrt x \left( {\sqrt x - 4} \right) + 2$ = 0

$\Rightarrow$ $\left( {\sqrt x - 2} \right)$ + $x - 4\sqrt x$ + 2 = 0

$\Rightarrow$ $x - 3\sqrt x$ = 0

$\Rightarrow$ $\sqrt x \left( {\sqrt x - 3} \right)$ = 0

$\therefore$ $\sqrt x$ = 0 or 3

$\sqrt x$ = 0 is not possible as $\sqrt x \ge 2$ .

So, $\sqrt x$ = 3

or $x$ = 9

Case 2 : When \sqrt x < 2

then $\left| {\sqrt x - 2} \right| =$ $- \left( {\sqrt x - 2} \right)$ = $2 - \sqrt x$

$\therefore$ The given equation becomes,

$\left( {2 - \sqrt x } \right)$ + $\sqrt x \left( {\sqrt x - 4} \right) + 2$ = 0

$\Rightarrow$ ${2 - \sqrt x }$ + $x - 4\sqrt x$ + 2 = 0

$\Rightarrow$ $x - 5\sqrt x + 4$ = 0

$\Rightarrow$ $x - 4\sqrt x - \sqrt x + 4$ = 0

$\Rightarrow$ $\sqrt x \left( {\sqrt x - 4} \right)$ $-\left( {\sqrt x - 4} \right)$ = 0

$\Rightarrow$ $\left( {\sqrt x - 4} \right)$ $\left( {\sqrt x - 1} \right)$ = 0

$\therefore$ $\sqrt x$ = 4 or 1

$\sqrt x$ = 4 is not possible as \sqrt x < 2 .

$\therefore$ $\sqrt x$ = 1

or $x$ = 1

So, Sum of all solutions = 9 + 1 = 10