Theory of Equations 1 Question 38
39. The sum of all the real roots of the equation $|x-2|^{2}+|x-2|-2=0$ is…… .
(1997, 2M)
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Solution:
- Given, $|x-2|^{2}+|x-2|-2=0$
Case I When $x \geq 2$
$$ \begin{array}{cc} \Rightarrow & (x-2)^{2}+(x-2)-2=0 \\ \Rightarrow & x^{2}+4-4 x+x-2-2=0 \\ \Rightarrow & x^{2}-3 x=0 \\ \Rightarrow & x(x-3)=0 \\ \Rightarrow & x=0,3 \\ \Rightarrow & x=3 \end{array} $$
Case II When $x<2$
$\Rightarrow \quad{-(x-2)}^{2}-(x-2)-2=0$
$\Rightarrow \quad(x-2)^{2}-x+2-2=0$
$\Rightarrow \quad x^{2}+4-4 x-x=0$
$\Rightarrow \quad x^{2}-4 x-(x-4)=0$
$\Rightarrow \quad x(x-4)-1(x-4)=0$
$\Rightarrow \quad(x-1)(x-4)=0$
$$ \begin{array}{ll} \Rightarrow & x=1,4 \\ \Rightarrow & x=1 \end{array} $$
[4 is rejected]
Hence, the sum of the roots is $3+1=4$.
Alternate Solution
Given, $|x-2|^{2}+|x-2|-2=0$
$\Rightarrow(|x-2|+2)(|x-2|-1)=0$
$\therefore \quad|x-2|=-2,1 \quad$ [neglecting -2 ]
$\Rightarrow \quad|x-2|=1 \quad \Rightarrow \quad x=3,1$
$\Rightarrow \quad$ Sum of the roots $=4$
