The numbers of pairs (a, b) of real numbers, such that whenever α is a root of the equation x² + ax

Q: Why do we only check fixed points and 2-cycles? Could there be a 3-cycle?

That is a brilliant observation! However, remember that our quadratic equation $x^2 + ax + b = 0$ has at most two roots. A 3-cycle would require the set of roots to contain at least three distinct values, which is impossible for a quadratic equation. We only need to consider mappings of a set with at most two elements.

Q: How do I know I have found all the cases? Is there a systematic way?

Yes, there is! We partition the problem based on the cardinality of the image of the set of roots. We checked the case where the roots are the same (repeated roots), where they are distinct and map to each other (a 2-cycle), and where they map to fixed points. This covers all logical possibilities for a set of size two.

Q: Wait, why did we reject $\alpha = \beta$ in the distinct roots case?

We didn't reject it; we handled it separately! In Case 1, we assumed $\alpha = \beta$ and found two pairs. In Case 2, we explicitly assumed $\alpha \neq \beta$ to find the remaining pairs. By separating the cases, we ensure we don't double-count or miss any solutions.

Question

The numbers of pairs $(a, b)$ of real numbers, such that whenever $\alpha$ is a root of the equation $x^2 + ax + b = 0$, $\alpha^2 - 2$ is also a root of this equation, is :

Options:

(A) 6

(B) 2

(C) 4

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The quadratic equation is given by $x^2 + ax + b = 0$ with roots $\alpha$ and $\beta$. We are given the constraint that if $\alpha$ is a root, then $\alpha^2 - 2$ must also be a root.

This implies that the set of roots $\{\alpha, \beta\}$ is invariant under the transformation $f(x) = x^2 - 2$. We seek to identify all pairs $(a, b)$ that satisfy this condition.

### Phase 1: The Fixed Points

The simplest invariant set occurs when the roots are fixed points of the transformation. We solve $f(x) = x$:

$$x^2 - 2 = x \Rightarrow x^2 - x - 2 = 0$$

Factoring the quadratic, we obtain $(x - 2)(x + 1) = 0$. Thus, the fixed points are $x = 2$ and $x = -1$.

If the roots are repeated ($\alpha = \beta$), then $\alpha$ must be a fixed point:

1. If $\alpha = \beta = -1$, then $\alpha + \beta = -2 = -a \Rightarrow a = 2$, and $\alpha\beta = 1 = b$. This yields the pair **$(2, 1)$**.

2. If $\alpha = \beta = 2$, then $\alpha + \beta = 4 = -a \Rightarrow a = -4$, and $\alpha\beta = 4 = b$. This yields the pair **$(-4, 4)$**.

### Phase 2: The 2-Cycle

We now consider the case where the roots are distinct and swap places, such that $f(\alpha) = \beta$ and $f(\beta) = \alpha$. This implies:

$$\alpha^2 - 2 = \beta \quad 	ext{and} \quad \beta^2 - 2 = \alpha$$

Subtracting these equations yields $\alpha^2 - \beta^2 = \beta - \alpha$. Since $\alpha 
eq \beta$, we divide by $(\alpha - \beta)$ to find:

$$\alpha + \beta = -1$$

Adding the two original equations gives $\alpha^2 + \beta^2 - 4 = \alpha + \beta$. Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$, we substitute the sum:

$$(-1)^2 - 2\alpha\beta - 4 = -1 \Rightarrow -2\alpha\beta = 2 \Rightarrow \alpha\beta = -1$$

With $\alpha + \beta = -1$ and $\alpha\beta = -1$, we find $a = -(\alpha + \beta) = 1$ and $b = \alpha\beta = -1$. This yields the pair **$(1, -1)$**.

### Phase 3: The Asymmetric Mapping

Finally, we consider cases where one root maps to the other, while the second root is a fixed point.

1. If $\beta = 2$ is a fixed point, then $\alpha^2 - 2 = 2 \Rightarrow \alpha = \pm 2$. Since $\alpha 
eq \beta$, we take $\alpha = -2$. The sum is $-2 + 2 = 0 = -a \Rightarrow a = 0$, and the product is $-4 = b$. This yields **$(0, -4)$**.

2. If $\beta = -1$ is a fixed point, then $\alpha^2 - 2 = -1 \Rightarrow \alpha = \pm 1$. Since $\alpha 
eq \beta$, we take $\alpha = 1$. The sum is $1 + (-1) = 0 = -a \Rightarrow a = 0$, and the product is $-1 = b$. This yields **$(0, -1)$**.

3. If the roots are the two distinct fixed points $\alpha = 2$ and $\beta = -1$, the sum is $1 = -a \Rightarrow a = -1$, and the product is $-2 = b$. This yields **$(-1, -2)$**.

### Final Summary

By exhaustively analyzing the dynamics of the transformation $f(x) = x^2 - 2$, we have identified the six valid pairs $(a, b)$:

**$(2, 1), (-4, 4), (-1, -2), (1, -1), (0, -4), 	ext{ and } (0, -1)$**.

Animated Solution for Mathematics - Quadratic Equations: The numbers of pairs $(a, b)$ of real numbers, such that whenever $α$ is a root of the equation $x^{2} + a x + b = 0$ , $α^{2} - 2$ is also a root of this equation, is :

Select Answer:

Visualized Solution

Defining the Problem

Case 1: Repeated Roots

Solving for Fixed Points

Finding $(a, b)$ for $α = - 1$

Finding $(a, b)$ for $α = 2$

Case 2: Distinct Roots

Subcase 2.1: Both are Fixed Points

Subcase 2.2: The 2-Cycle

Solving for the Product in 2-Cycle

Finding $(a, b)$ for the 2-Cycle

Subcase 2.3: Mapping to a Fixed Point

Finding $(a, b)$ for Roots ${- 2, 2}$

Mapping to the other Fixed Point

Finding $(a, b)$ for Roots ${- 1, 1}$

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Summary

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