For the equation

Question:

For the equation $|x|^{2}+|x|-6=0$, the sum of the real roots is

(a) 1

(b) 0

(c) 2

(d) none of these

Solution:

(b) 0

Let $p=|x|$

$\Rightarrow p^{2}+p-6=0$

$\Rightarrow p^{2}+3 p-2 p-6=0$

$\Rightarrow(p+3)(p-2)=0$

$\Rightarrow p=-3,2$

Also, $|x|=p$

$\Rightarrow|x|=2$, or $|x|=-3$

Modulus can not be negative,$\therefore|x|=2$

$\Rightarrow x=\pm 2$

$\Rightarrow x=2$ or $-2$

Sum of the roots of $x$ is 0

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