Let \(a, b, c, p, q\) be real numbers.
Suppose, \(\alpha, \beta\) are the roots of the equation \(x^2+2 p x+q=0\) and \(\alpha, \frac{1}{\beta}\) are the roots of the equation \(a x^2+2 b x+c=0\), where \(\beta^2 \notin\{-1,0,1\}\).
Statement \(1\left(p^2-q\right)\left(b^2-a c\right) \geq 0\).
Statement \(2 b \neq p a\) or \(c \neq q a\).

Given, \(\alpha\) and \(\beta\) are the roots of \(x^2+2 p x+q=0\).
\(\therefore \alpha+\beta =-2 p \)
\( \text {and } \alpha \beta =q\)
\(\alpha\) 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
\(
\frac{\alpha}{\beta}=\frac{c}{a}
\)
\( \text {Now, } \left(p^2-q\right)\left(b^2-a c\right) =\left[\left(\frac{\alpha+\beta}{-2}\right)^2-\alpha \beta\right]\)\(\left[\left(\frac{\alpha+\frac{1}{\beta}}{2}\right)^2-\frac{\alpha}{\beta}\right] a^2 \)
\( =\frac{(\alpha-\beta)^2}{16}\left(\alpha-\frac{1}{\beta}\right)^2 \cdot a^2 \geq 0\)
Statement 1 is true.
\(\text {Again now, } p a=-\left(\frac{\alpha+\beta}{2}\right) a=-\frac{a}{2}(\alpha+\beta) \)
\( \text {and } b=-\frac{a}{2}\left(\alpha+\frac{1}{\beta}\right) \)
\( p a \neq b \Rightarrow \alpha+\frac{1}{\beta} \neq \alpha+\beta \)
\( \Rightarrow \beta \neq 1 \)
\( \because \beta^2 \neq\{-1,0,1\}, \text {correct. } \)
\( \text {Similarly, if } c \neq q x \)
\( \Rightarrow a \frac{\alpha}{\beta} \neq a \alpha \beta \Rightarrow \alpha\left(\beta-\frac{1}{\beta}\right) \neq 0\)
\(\Rightarrow \alpha \neq 0 \text { and } \beta-\frac{1}{\beta} \neq 0 \)
\( \Rightarrow \beta \neq\{-1,0,1\}\)
Statement 2 is true.
Both Statement 1 and Statement 2 are true. But Statement 2 does not explain Statement 1.