Evaluate Statements on Discriminants and Coefficients of Quadratic Equations

To evaluate the given statements, let us analyze the properties of the roots and coefficients of the two quadratic equations step-by-step.

Analysis of STATEMENT-1:

We are given two quadratic equations with real coefficients $a, b, c, p, q$:

1. $x^2 + 2px + q = 0$ with roots $\alpha, \beta$.
2. $ax^2 + 2bx + c = 0$ with roots $\alpha, \frac{1}{\beta}$.

From Vieta's formulas for the second equation, the sum of its roots is:
$\alpha + \frac{1}{\beta} = -\frac{2b}{a}$

Since $a, b \in \mathbb{R}$ with $a \neq 0$, the sum $-\frac{2b}{a}$ must be a real number.

Now, let us determine whether the roots $\alpha, \beta$ can be non-real complex numbers:

• If $\alpha$ and $\beta$ were non-real complex numbers, then because $x^2 + 2px + q = 0$ has real coefficients $p$ and $q$, the roots must be complex conjugates of each other, i.e., $\beta = \bar{\alpha}$.
• Substituting $\beta = \bar{\alpha}$ into the sum of roots of the second equation gives:
  $\alpha + \frac{1}{\beta} = \alpha + \frac{1}{\bar{\alpha}} = \alpha + \frac{\alpha}{|\alpha|^2} = \alpha \left(1 + \frac{1}{|\alpha|^2}\right)$
• Since $|\alpha|^2 > 0$, the term $\left(1 + \frac{1}{|\alpha|^2}\right)$ is a positive real number.
• For $\alpha \left(1 + \frac{1}{|\alpha|^2}\right)$ to be a real number, $\alpha$ itself must be a real number.

This contradicts the assumption that $\alpha$ is non-real. Thus, $\alpha$ and $\beta$ must be real numbers.

Since $\alpha$ and $\beta$ are real numbers:

1. The first equation $x^2 + 2px + q = 0$ has real roots, so its discriminant must be non-negative:
   $D_1 = 4(p^2 - q) \ge 0 \implies p^2 - q \ge 0$
2. The roots of the second equation are $\alpha$ and $\frac{1}{\beta}$ (where $\beta \neq 0$ since $\beta^2 \notin \{-1, 0, 1\}$). Since both roots are real, its discriminant must also be non-negative:
   $D_2 = 4(b^2 - ac) \ge 0 \implies b^2 - ac \ge 0$

Multiplying these two non-negative terms yields:
$(p^2 - q)(b^2 - ac) \ge 0$

Thus, STATEMENT-1 is True.

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Analysis of STATEMENT-2:

Suppose $b = pa$ and $c = qa$.
Substituting these into the second equation $ax^2 + 2bx + c = 0$ gives:
$a(x^2 + 2px + q) = 0$

Since $a \neq 0$, this implies that $ax^2 + 2bx + c = 0$ has the exact same roots (with multiplicity) as $x^2 + 2px + q = 0$.

• The multiset of roots of the first equation is $\{\alpha, \beta\}$.
• The multiset of roots of the second equation is $\left\{\alpha, \frac{1}{\beta}\right\}$.

Equating the two multisets gives two possibilities:

1. $\beta = \frac{1}{\beta} \implies \beta^2 = 1$
2. $\alpha = \frac{1}{\beta}$ and $\beta = \alpha \implies \beta = \frac{1}{\beta} \implies \beta^2 = 1$

Both cases imply $\beta^2 = 1$, which directly contradicts the given condition that $\beta^2 \notin \{-1, 0, 1\}$.

Hence, it is impossible to have both $b = pa$ and $c = qa$ simultaneously. Therefore, at least one of these conditions must fail:
$b \neq pa \quad \text{or} \quad c \neq qa$

Thus, STATEMENT-2 is True.

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Conclusion:

While both statements are logically true:

• STATEMENT-1 is true because the roots $\alpha$ and $\beta$ are real numbers, making both discriminants non-negative independently.
• STATEMENT-2 is true due to the constraint $\beta^2 \notin \{-1, 0, 1\}$, but it does not serve as the explanation for the non-negativity of the product of the discriminants in STATEMENT-1.

Therefore, STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1.

Correct Answer: B