
6. Find all real numbers $a$ for which the equation

$$
x^{2}+(a-2) x+1=3|x|
$$

has exactly three distinct real solutions in $x$.


Solution: If $x \geq 0$, then the given equation assumes the form,

$$
x^{2}+(a-5) x+1=0
$$

If $x4$ and $(a+1)^{2}=4$;

(II) or $(a-5)^{2}=4$ and $(a+1)^{2}>4$.

Case (I) From $(a+1)^{2}=4$, we have $a=1$ or -3 . But only $a=1$ satisfies $(a-5)^{2}>4$. Thus $a=1$. Also when $a=1$, equation (1) has solutions $x=2+\sqrt{3}$; and (2) has solutions $x=-1,-1$. As $2 \pm \sqrt{3}>0$ and $-14$. When $a=3$, equation (1) has solutions $x=1,1$ and (2) has the solutions $x=-2 \pm \sqrt{3}$. As $1>0$ and $-2 \pm \sqrt{3}<0$, we see that $a=3$ is in fact a solution.

When $a=7$, equation (1) has solutions $x=-1,-1$, which are negative contradicting $x \geq 0$.

Thus $a=1, a=3$ are the two desired values.
