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

Given,

$x\left|x\right| - 5$\left|x + 2\right| + 6 = 0$</p><p>Now taking case \left(1\right)$ when $x < - 2$ we get,

$- {x}^{2} + 5x + 10 + 6 = 0$

$⇒ {x}^{2} - 5x - 16 = 0$

$⇒ x = \dfrac{5 - \sqrt{89}}{2}$, ignoring $x = \dfrac{5 + \sqrt{89}}{2}$ as $x < - 2$, so one solution is possible here,

Now taking case \left(2\right)$when$- 2 \le x < 0$we get,</p><p>$- {x}^{2} - 5x - 10 + 6 = 0$</p><p>$⇒ {x}^{2} + 5x + 4 = 0$</p><p>$⇒ x = - 1$, ignoring$x = - 4$as$- 2 \le x < 0$</p><p>Now taking case \left(3\right)$, when $x \ge 0$ we get,

${x}^{2} - 5x - 10 + 6 = 0$

$⇒ {x}^{2} - 5x - 4 = 0$

$⇒ x = \dfrac{5 + \sqrt{41}}{2}$, ignoring $x = \dfrac{5 - \sqrt{41}}{2}$ as $x \ge 0$

So, from all cases we get total $3$ solutions.