Question

If \(P(x) = ax^2 + bx + c\) and \(Q(x) = -ax^2 + dx + c\) where \(ac \neq 0\), then \(P(x) \cdot Q(x) = 0\) has:

• A. 2 real roots
• B. at least two real roots
• C. 4 real roots
• D. no real roots

Hint:

When subtracting quadratic equations, always simplify and factor the terms to identify the real roots. In this case, factoring helped us identify the roots clearly.

Answer 1

The Correct Option is B

Step 1: We start with the equations \( P(x) = ax^2 + bx + c \) and \( Q(x) = -ax^2 + dx + c \). The goal is to solve \( P(x) \cdot Q(x) = 0 \).

Step 2: Calculate the difference between \( P(x) \) and \( Q(x) \):

\[ P(x) \cdot Q(x) = (ax^2 + bx + c) - (-ax^2 + dx + c). \]

Simplify:

\[ P(x) \cdot Q(x) = ax^2 + bx + c + ax^2 - dx - c = 2ax^2 + (b - d)x. \]

Step 3: The resulting equation is:

\[ 2ax^2 + (b - d)x = 0. \]

Step 4: Factor out \( x \):

\[ x(2ax + b - d) = 0. \]

Step 5: The solutions are \( x = 0 \) or \( 2ax + b - d = 0 \). Solving the second equation gives:

\[ x = \frac{d - b}{2a}. \]

Thus, the equation has two real roots: \( x = 0 \) and \( x = \frac{d - b}{2a} \).