Answer

Solution

1. 1

   Case 1: $x\ge 0$. $x^2+1=3x\Rightarrow x^2-3x+1=0\Rightarrow x=\dfrac{3\pm\sqrt 5}{2}>0$. Both positive (since $\frac{3-\sqrt 5}{2}\approx 0.38>0$). 2 solutions.

2. 2

   Case 2: $x<0$. $x|x|=-x^2$, $|x|=-x$. $-x^2+1=-3x\Rightarrow x^2-3x-1=0\Rightarrow x=\dfrac{3\pm\sqrt{13}}{2}$. Need $x<0$: $\dfrac{3-\sqrt{13}}{2}\approx-0.30<0$; $\dfrac{3+\sqrt{13}}{2}>0$ (reject). 1 solution.

3. 3

   Total: $3$ real solutions.

Answer structure / marking notes

Content note