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4-2.Quadratic Equations and Inequations
hard
The sum of the solutions of the equation $\left| {\sqrt x - 2} \right| + \sqrt x \left( {\sqrt x - 4} \right) + 2 = 0\left( {x > 0} \right)$ is equal to
A
$9$
B
$4$
C
$10$
D
$12$
(JEE MAIN-2019)
Solution
$|\sqrt{x}-2|+\sqrt{x}(\sqrt{x}-4)+2=0$
$|\sqrt{x}-2|+(\sqrt{x})^{2}-4 \sqrt{x}+2=0$
$|\sqrt{x}-2|^{2}+|\sqrt{x}-2|-2=0$
$|\sqrt{x}-2|=-2(\text { not possible })$ or $|\sqrt{x}-2|=1$
$\sqrt{x}-2=1,-1$
$\sqrt{x}=3,1$
$x=9,1$
Sum $=10$
Standard 11
Mathematics
