The number of the real solutions of the equation: x|x+3|+|x-1|-2=0 is

Question

The number of the real solutions of the equation: $x|x+3|+|x-1|-2=0$ is

Accepted Answer

given $x|x+3|+|x-1|-2=0$.

critical points $x=-3,1$.

two critical points means 3 cases.

Case 1 : $x$$$\begin{gathered} -x(x+3)-(x-1)-2=0 \ -{x}^{2}-3x-x+1-2=0 \ -{x}^{2}-4x-1=0\implies {x}^{2}+4x+1=0 \ x=\frac{-4\pm \sqrt{16-4}}{2}=\frac{-4\pm 2\sqrt{3}}{2}=-2\pm \sqrt{3} \end{gathered}$$$

since $xCase 2 : $-3\le x\le 1,|x+3|=x+3$ and $|x-1|=-(x-1)$. equation become

$\begin{gathered} x(x+3)-(x-1)-2=0. \ {x}^{2}+3x-x+1-2=0. \ {x}^{2}+2x-1=0. \ x=\frac{-2\pm \sqrt{4+4}}{2}=\frac{-2\pm 2\sqrt{2}}{2}=-1\pm \sqrt{2}. \end{gathered}$

since $-3\le x\le 1$ so $x=-1+\sqrt{2},-1-\sqrt{2}$ both accepted.

Case 3 : $x>1,|x+3|=x+3|x-1|=x-1$

equation become

$x(x+3)+x-1-2=0.$

${x}^{2}+4x-3=0.$

$x=\frac{-4\pm \sqrt{16+12}}{2}=\frac{-4\pm 2\sqrt{7}}{2}=-2\pm \sqrt{7}$. Checking positive solution. since $x>1$, so $x=\sqrt{7}-2\approx 0.645$, so no solution.

so final solution is

$x=-2-\sqrt{3},-1+\sqrt{2},-1-\sqrt{2}.$

Number of solution = 3

Solutions · 1