Question

Let $m$ and $p$ be real numbers such that the polynomial $f(x) = x^2 + mx + p$ has two distinct negative rational roots. Then the polynomial $g(x) = x^2 - (m^2 - 2p)x + p^2$ has distinct

• A. positive rational roots
• B. positive irrational roots
• C. negative rational roots
• D. negative irrational roots

Hint:

Using a simple example can save time. Let the roots of $f(x)$ be $-1$ and $-2$.
Then $f(x) = (x+1)(x+2) = x^2 + 3x + 2 \implies m = 3, p = 2$.
Substituting these into $g(x)$ gives $g(x) = x^2 - (9 - 4)x + 4 = x^2 - 5x + 4 = (x-1)(x-4)$.
The roots of $g(x)$ are 1 and 4, which are distinct, positive, and rational.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given information about the roots of a quadratic polynomial $f(x)$ (distinct, negative, and rational).
We need to determine the nature of the roots (distinctness, sign, and rationality) of a related quadratic polynomial $g(x)$.

Step 2: Key Formula or Approach:
For a quadratic equation $ax^2 + bx + c = 0$:
- The discriminant is $D = b^2 - 4ac$.
- The roots are rational if and only if $D$ is a perfect square of a rational number (assuming coefficients are rational).
- Vieta's formulas state that the sum of roots is $-b/a$ and the product of roots is $c/a$.

Step 3: Detailed Explanation:

• Let $\alpha$ and $\beta$ be the roots of $f(x) = x^2 + mx + p$.
Since $\alpha$ and $\beta$ are rational and distinct, the coefficients $m$ and $p$ must be rational numbers.
Since the roots are negative, we have $\alpha \lt 0$ and $\beta \lt 0$.

• Using Vieta's formulas:
\[ m = -(\alpha + \beta) \gt 0 \]
\[ p = \alpha\beta \gt 0 \]

• Since the roots of $f(x)$ are distinct and rational, the discriminant $D_f$ must be a positive perfect square of a rational number, say $k^2$ ($k \in \mathbb{Q}, k \gt 0$):
\[ D_f = m^2 - 4p = k^2 \gt 0 \]

• Now let us examine the polynomial $g(x) = x^2 - (m^2 - 2p)x + p^2$.
Let us compute its discriminant $D_g$:
\[ D_g = (m^2 - 2p)^2 - 4p^2 \]
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
\[ D_g = (m^2 - 2p - 2p)(m^2 - 2p + 2p) \]
\[ D_g = (m^2 - 4p)(m^2) \]

• Substituting $m^2 - 4p = k^2$ into the expression:
\[ D_g = k^2 m^2 = (km)^2 \]
Since $k$ and $m$ are rational numbers, their product $km$ is also rational. Thus, $D_g$ is a perfect square of a rational number.
Since $k \gt 0$ and $m \gt 0$, we have $D_g \gt 0$.
Therefore, the roots of $g(x)$ are real, distinct, and rational.

• Let us analyze the signs of the roots of $g(x)$ using Vieta's formulas:
- Product of roots:
\[ \text{Product} = p^2 \gt 0 \]
- Sum of roots:
\[ \text{Sum} = m^2 - 2p \]
Substitute $m = -(\alpha + \beta)$ and $p = \alpha\beta$:
\[ \text{Sum} = (\alpha + \beta)^2 - 2\alpha\beta = \alpha^2 + \beta^2 \]
Since $\alpha \neq \beta$ and both are non-zero, $\alpha^2 + \beta^2 \gt 0$.

• Since both the sum and the product of the roots are positive, both roots of $g(x)$ must be positive.



Step 4: Final Answer:
The polynomial $g(x)$ has distinct positive rational roots.