Theory of Equations 5 Question 3
3. The sum of the solutions of the equation $|\sqrt{x}-2|+\sqrt{x}(\sqrt{x}-4)+2=0(x>0)$ is equal to
(2019 Main, 8 April I)
(a) 9
(b) 12
(c) 4
(d) 10
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Solution:
Key Idea Reduce the given equation into quadratic equation.
Given equation is
$$ |\sqrt{x}-2|+\sqrt{x}(\sqrt{x}-4)+2=0 $$
$\Rightarrow|\sqrt{x}-2|+x-4 \sqrt{x}+4=2$
$\Rightarrow|\sqrt{x}-2|+(\sqrt{x}-2)^{2}=2$
$\Rightarrow(|\sqrt{x}-2|)^{2}+|\sqrt{x}-2|-2=0$
Let $|\sqrt{x}-2|=y$, then above equation reduced to
$$ y^{2}+y-2=0 \Rightarrow y^{2}+2 y-y-2=0 $$
$\Rightarrow y(y+2)-1(y+2)=0 \Rightarrow(y+2)(y-1)=0$
$\Rightarrow \quad y=1,-2$
$$ \therefore \quad-y=1 \quad[\because y=|\sqrt{x}-2| \geq 0] $$
$$ \Rightarrow \quad|\sqrt{x}-2|=1 $$
$\Rightarrow \quad \sqrt{x}-2= \pm 1$
$\Rightarrow \quad \sqrt{x}=3$ or 1
$$ \Rightarrow \quad x=9 \text { or } 1 $$
$\therefore$ Sum of roots $=9+1=10$
