
2. Find the set of all real values of $a$ for which the real polynomial equation $P(x)=x^{2}-2 a x+b=0$ has real roots given that $P(0) \cdot P(1) \cdot P(2) \neq 0$ and $P(0), P(1), P(2)$ form a geometric progression.


Solution: Since none of $P(0), P(1), P(2)$ is zero, we have $P(1)^{2}=P(0) P(2)$. Observe that $P(0)=b, P(1)=1-2 a+b$ and $P(2)=4-4 a+b$. Therefore we obtain $(1-2 a+b)^{2}=b(4-4 a+b)$. This simplifies to $(2 a-1)^{2}=2 b$. Now $P(x)=0$ has real roots if and only if $4 a^{2} \geq 4 b$. Hence $2 a^{2} \geq 2 b=(2 a-1)^{2}$. This reduces to $2 a^{2}-4 a+1 \leq 0$. Consider the quadratic equation $2 X^{2}-4 X+1=0$. This has two roots

$$
\alpha=\frac{2+\sqrt{2}}{2}, \quad \beta=\frac{2-\sqrt{2}}{2}
$$

Hence $2 a^{2}-4 a+1 \leq 0$ if and only if $a$ lies in the interval $\left[\frac{2-\sqrt{2}}{2}, \frac{2+\sqrt{2}}{2}\right]$.
