Let $a, b, c, p, q$ be real numbers. Suppose, $\alpha, \beta$ are the roots of the equation $x^2+2 p x+q=0$…
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 $c \neq q a$.
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
Statement 1 is true, Statement 2 is false.
Statement 1 is false, Statement 2 is true
Solution
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 \cdot a^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 \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
\end{array}
$
$
\begin{array}{ll}
\Rightarrow & \alpha \neq 0 \text { and } \beta-\frac{1}{\beta} \neq 0 \\
\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