Question

The number of real roots of the equation $x^2+3\left|x\right|+2=0$ is ________.

Answer 1

Given: $x^2+3\left|x\right|+2=0$

Since,
$$\begin{gathered} \left|x\right|=\left\{\begin{array}{l}x,\mathrm{if}x\ge 0\\ -x,\mathrm{if}x<0\end{array}\right. \\ \mathrm{Thus}, \\ {x}^2+3\left|x\right|+2=0\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as} \\ \left\{\begin{array}{l}{x}^2+3x+2=0,\mathrm{if}x\ge 0...\left(1\right)\\ x^2-3x+2=0,\mathrm{if}x<0...\left(2\right)\end{array}\right. \\ \mathrm{Solving}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{if}x\ge 0. \\ {x}^2+3x+2=0 \\ \Rightarrow x^2+2x+x+2=0 \\ \Rightarrow x\left(x+2\right)+1\left(x+2\right)=0 \\ \Rightarrow \left(x+1\right)\left(x+2\right)=0 \\ \Rightarrow x=-1,-2 \\ \mathrm{But},x\ge 0 \\ \therefore x\ne -1,-2 \\ \mathrm{Solving}\left(2\right),\mathrm{we}\mathrm{get} \\ \mathrm{if}x<0. \\ {x}^2-3x+2=0 \\ \Rightarrow x^2-2x-x+2=0 \\ \Rightarrow x\left(x-2\right)-1\left(x-2\right)=0 \\ \Rightarrow \left(x-1\right)\left(x-2\right)=0 \\ \Rightarrow x=1,2 \\ \mathrm{But},x<0 \\ \therefore x\ne 1,2 \end{gathered}$$

Hence, the number of real roots of the equation $x^2+3\left|x\right|+2=0$ is 0.