Question

The number of real roots of the equation x² + 5 |x| + 4 = 0 is _________.

Answer 1

$$\begin{gathered} x^2+5\left|x\right|+4=0 \\ \mathrm{for}\left|x\right|,\mathrm{we}\mathrm{have}2\mathrm{cases} \\ \mathrm{case}\left(\mathrm{i}\right) \\ \mathrm{if}x>0 \\ \mathrm{Quadratic}\mathrm{equation}\mathrm{is}{x}^2+5x+4=0 \\ \mathrm{i}.\mathrm{e}.x^2+4x+x+4=0 \\ \mathrm{i}.\mathrm{e}.\left(x+4\right)\left(x+1\right)=0 \\ \mathrm{i}.\mathrm{e}.x=-1\mathrm{or}-4 \\ \mathrm{which}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{since}\mathit{}x>0 \\ \mathrm{case}\left(\mathrm{ii}\right)\mathrm{for}x<0, \\ \mathrm{Quadratic}\mathrm{equation}\mathrm{is}{x}^2-5x+4=0 \\ \mathrm{i}.\mathrm{e}.x^2-4x-x+4=0 \\ \mathrm{i}.\mathrm{e}.\left(x-4\right)\left(x-1\right)=0 \\ \mathrm{i}.\mathrm{e}.x=1\mathrm{or}x=4 \\ \mathrm{which}\mathrm{is}\mathrm{not}\mathrm{possible},\mathrm{since}x\mathrm{is}\mathrm{negative}\mathrm{in}\mathrm{this}\mathrm{case} \\ \mathrm{Hence},\mathrm{no}\mathrm{real}\mathrm{solution}\mathrm{exist}\mathrm{for}{x}^2+5\left|x\right|+4=0 \end{gathered}$$