Key Concepts and Formulas

• Sum and Product of Roots: For a quadratic equation $x^2 + ax + b = 0$, the sum of the roots is $\alpha + \beta = -a$, and the product of the roots is $\alpha\beta = b$.
• Quadratic Formula: The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• The problem states that if $\alpha$ is a root, then $\alpha^2 - 2$ is also a root.

Step-by-Step Solution

Step 1: Define the Problem and Roots

We are given the quadratic equation $x^2 + ax + b = 0$, where $a$ and $b$ are real numbers. Let $\alpha$ and $\beta$ be the roots of this equation. The problem states that if $\alpha$ is a root, then $\alpha^2 - 2$ is also a root. Our goal is to find the number of possible pairs $(a, b)$ that satisfy this condition.

Step 2: Consider the Case of Repeated Roots ($\alpha = \beta$)

If the quadratic equation has a repeated root, then $\alpha = \beta$. In this case, the condition states that $\alpha^2 - 2 = \alpha$. Why this step? We start with the simplest case to find some initial solutions.

$\alpha^2 - \alpha - 2 = 0$ $(\alpha - 2)(\alpha + 1) = 0$ This gives two possible values for $\alpha$: $\alpha = 2$ or $\alpha = -1$.

Step 3: Find (a, b) for $\alpha = 2$

If the repeated root is $2$, then the quadratic equation is $(x - 2)^2 = 0$. Why this step? We want to find the corresponding quadratic equation and thus the values of a and b. $x^2 - 4x + 4 = 0$ Comparing with $x^2 + ax + b = 0$, we have $a = -4$ and $b = 4$. So, $(a, b) = (-4, 4)$.

Step 4: Find (a, b) for $\alpha = -1$

If the repeated root is $-1$, then the quadratic equation is $(x + 1)^2 = 0$. Why this step? Similar to step 3, we find the a and b values. $x^2 + 2x + 1 = 0$ Comparing with $x^2 + ax + b = 0$, we have $a = 2$ and $b = 1$. So, $(a, b) = (2, 1)$.

Step 5: Consider the Case of Distinct Roots ($\alpha \ne \beta$)

If the quadratic equation has two distinct roots, then $\alpha \ne \beta$. We know that if $\alpha$ is a root, then $\alpha^2 - 2$ is also a root. This gives us two possibilities:

1. $\alpha^2 - 2 = \beta$ and $\beta^2 - 2 = \alpha$ (roots are interchanged).
2. $\alpha^2 - 2 = \alpha$ and $\beta^2 - 2 = \beta$ (both roots are fixed points).

Why this step? Now we consider the more general case where the two roots may be different. The condition given by the problem implies these two possibilities.

Step 6: Analyze the case where $\alpha^2 - 2 = \beta$ and $\beta^2 - 2 = \alpha$

Subtracting the two equations, we get: $\alpha^2 - \beta^2 = \beta - \alpha$ $(\alpha - \beta)(\alpha + \beta) = \beta - \alpha$ $(\alpha - \beta)(\alpha + \beta) + (\alpha - \beta) = 0$ $(\alpha - \beta)(\alpha + \beta + 1) = 0$ Since $\alpha \ne \beta$, we have $\alpha + \beta + 1 = 0$, which implies $\alpha + \beta = -1$. Why this step? Here we perform algebraic manipulation to arrive at a simpler relation between alpha and beta.

Step 7: Solve for $\alpha$ and $\beta$

Substituting $\beta = -1 - \alpha$ into $\alpha^2 - 2 = \beta$, we have: $\alpha^2 - 2 = -1 - \alpha$ $\alpha^2 + \alpha - 1 = 0$ Using the quadratic formula, we find: $\alpha = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$ So, the roots are $\alpha = \frac{-1 + \sqrt{5}}{2}$ and $\beta = \frac{-1 - \sqrt{5}}{2}$. Then, $\alpha + \beta = -1$ and $\alpha\beta = \frac{(-1)^2 - (\sqrt{5})^2}{4} = \frac{1 - 5}{4} = -1$.

Step 8: Find (a, b) for $\alpha = \frac{-1 + \sqrt{5}}{2}$ and $\beta = \frac{-1 - \sqrt{5}}{2}$

Since $\alpha + \beta = -a$ and $\alpha\beta = b$, we have $a = 1$ and $b = -1$. So, $(a, b) = (1, -1)$.

Step 9: Analyze the case where $\alpha^2 - 2 = \alpha$ and $\beta^2 - 2 = \beta$

From Step 2, we know that the roots are $2$ and $-1$. Why this step? The fixed points of the function x^2 - 2 are the roots found in step 2.

Step 10: Find (a, b) for $\alpha = 2$ and $\beta = -1$

Then, $\alpha + \beta = 2 + (-1) = 1$ and $\alpha\beta = 2(-1) = -2$. So, $-a = 1 \implies a = -1$ and $b = -2$. Therefore, $(a, b) = (-1, -2)$.

Step 11: Summarize the Pairs (a, b)

The pairs $(a, b)$ are:

1. $(-4, 4)$
2. $(2, 1)$
3. $(1, -1)$
4. $(-1, -2)$

Therefore, there are 4 possible pairs $(a, b)$.

Common Mistakes & Tips

• Missing the Repeated Root Case: Forgetting to consider the case where the quadratic has a repeated root.
• Incorrectly Applying the Given Condition: Not fully understanding the implications of "whenever $\alpha$ is a root, $\alpha^2 - 2$ is also a root."
• Algebraic Errors: Making mistakes when solving the equations for $\alpha$ and $\beta$.

Summary

By systematically analyzing the cases of repeated and distinct roots, and by carefully applying the given condition, we found four distinct pairs of real numbers $(a, b)$ that satisfy the condition: $(-4, 4)$, $(2, 1)$, $(1, -1)$, and $(-1, -2)$.

The final answer is $\boxed{4}$, which corresponds to option (C).