The number of real roots of the equation x|x|-5|x+2|+6=0, is :

Question

The number of real roots of the equation $x|x|-5|x+2|+6=0$, is :

Accepted Answer

$$
\begin{aligned}
& x|x|-5|x+2|+6=0 \\
& \text { Case-I : } \\
& \text { When } xCase-II :

\end{aligned}$$

\& \text { When } -2 \leq x Case-III :

$$
\begin{aligned}
& \text { When } x \geq 0 \text { then } \\
& x^2-5(x+2)+6=0 \\
\Rightarrow & x^2-5 x-4=0 \\
& x=\frac{5 \pm \sqrt{25+16}}{2} \\
& =\frac{5 \pm \sqrt{41}}{2} \\
& x=\frac{5 - \sqrt{41}}{2} \text { is accepted } \\
\therefore  & 3 \text { real roots are possible. }
\end{aligned}
$$

Solutions · 1