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

Show Answer

Answer:

Correct Answer: 3. (d)

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 $=9+1=10$



NCERT Chapter Video Solution

Dual Pane