Mathematics - Quadratic Equation Question with Solution | TestHub

Given,

$x\left|x\right|-5\left|x+2\right|+6=0$

Now taking case $\left(1\right)$ when $x<-2$ we get,

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

$\Rightarrow x^2-5x-16=0$

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

Now taking case $\left(2\right)$ when $-2\le x<0$ we get,

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

$\Rightarrow x^2+5x+4=0$

$\Rightarrow x=-1$, ignoring $x=-4$ as $-2\le x<0$

Now taking case $\left(3\right)$, when $x\ge 0$ we get,

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

$\Rightarrow x^2-5x-4=0$

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

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