Question

For the equation ${\left|x\right|}^2+\left|x\right|-6=0$, the sum of the real roots is
(a) 1
(b) 0
(c) 2
(d) none of these

Answer 1

(b) 0

$$\begin{gathered} \mathrm{Let}p=\left|x\right| \\ \Rightarrow p^2+p-6=0 \\ \Rightarrow p^2+3p-2p-6=0 \\ \Rightarrow \left(p+3\right)\left(p-2\right)=0 \\ \Rightarrow p=-3,2 \\ \mathrm{Also},\left|x\right|=p \\ \Rightarrow \left|x\right|=2,\mathrm{or}\left|x\right|=-3 \\ \mathrm{Modulus}\mathrm{can}\mathrm{not}\mathrm{be}\mathrm{negative}, \\ \therefore \left|x\right|=2 \\ \Rightarrow x=\pm 2 \\ \Rightarrow x=2\mathrm{or}-2 \\ \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{roots}\mathrm{of}x\mathrm{is}0 \end{gathered}$$