Let a, b, c, p, q be real numbers. Suppose, α, β are the roots of...

Given, $\alpha$ and $\beta$ are the roots of $x^2+2 p x+q=0$. $$\begin{array}{rlrl}& \therefore \alpha+\beta& =-2 p \\ \text{ and } \alpha \beta& =q \end{array}$$\alpha$and$\frac{1}{\beta}$are the roots of$a x^2+2 b x+c=0$and$\alpha+\frac{1}{\beta}=\frac{-2 b}{a}$and$\frac{\alpha}{\beta}=\frac{c}{a} \text{ Now, } $$\begin{aligned} \left(p^2-q\right)\left(b^2-a c\right) & =\left[\left(\frac{\alpha+\beta}{-2}\right)^2-\alpha \beta\right] \left[\left(\frac{\alpha+\frac{1}{\beta}}{2}\right)^2-\frac{\alpha}{\beta}\right]a^2 \\ & =\frac{(\alpha-\beta)^2}{16}\left(\alpha-\frac{1}{\beta}\right)^2 \cdota^2 \geq 0 \end{aligned}$$ Statement 1 is true. $$\begin{array}{ll} \text{ Again now, }& p a=-\left(\frac{\alpha+\beta}{2}\right)a=-\frac{a}{2}(\alpha+\beta)\\ \text{ and }& b=-\frac{a}{2} \left(\alpha+\frac{1}{\beta}\right)\\ & p a ≠ b \Rightarrow \alpha+\frac{1}{\beta} \neq \alpha+\beta\\ \Rightarrow & \beta \neq1 \\ \because & \beta^2≠ \{-1,0,1\}, \text{ correct. }\\ \text{ Similarly, if }& c ≠ q x \\ \Rightarrow & a \frac{\alpha}{\beta} \neqa \alpha \beta \Rightarrow \alpha\left(\beta-\frac{1}{\beta}\right) \neq0 \end{array}$$ $$\begin{array}{ll} \Rightarrow& \alpha \neq0 \text{ and } \ beta-\frac{1}{\beta} \neq0 \\ \Rightarrow & \beta \neq\{-1,0,1\} \end{array}$$ Statement 2 is true. Both Statement 1 and Statement 2 are true. But Statement 2 does not explain Statement 1.$