Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$. Let the sets
$\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of hyperbolas}\} = (\alpha, \beta)$, and
$\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of an ellipse and a hyperbola, respectively}\} = (\gamma, \infty)$.
Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:
Understanding the Roots and Eccentricities
The given equation is $x^2 - ax + 2 = 0$. Let $e_1$ and $e_2$ be its distinct roots.
From Vieta's formulas:
- Sum of roots: $e_1 + e_2 = a$
- Product of roots: $e_1 e_2 = 2$
For distinct real roots, the discriminant must be positive: $D = a^2 - 4(1)(2) = a^2 - 8 > 0$. This implies $a^2 > 8$, so $a \in (-\infty, -2\sqrt{2}) \cup (2\sqrt{2}, \infty)$.
Finding the Interval $(\alpha, \beta)$
This set corresponds to $e_1$ and $e_2$ being eccentricities of hyperbolas. Thus, $e_1 > 1$ and $e_2 > 1$.
Consider the quadratic function $f(t) = t^2 - at + 2$. For both roots to be greater than 1, the following conditions must hold:
- Discriminant $D > 0 \implies a^2 > 8$.
- Sum of roots $e_1 + e_2 = a > 1 + 1 = 2$.
- $f(1) > 0$. $f(1) = 1^2 - a(1) + 2 = 3 - a$. So, $3 - a > 0 \implies a < 3$.
Combining $a^2 > 8$, $a > 2$, and $a < 3$: Since $2\sqrt{2} \approx 2.828$, the condition $a^2 > 8$ means $a > 2\sqrt{2}$ or $a < -2\sqrt{2}$. The intersection of $(a > 2\sqrt{2} \lor a < -2\sqrt{2})$, $a > 2$, and $a < 3$ is $(2\sqrt{2}, 3)$.
Therefore, $(\alpha, \beta) = (2\sqrt{2}, 3)$. So, $\alpha = 2\sqrt{2}$ and $\beta = 3$.
Finding the Interval $(\gamma, \infty)$
This set corresponds to $e_1$ being the eccentricity of an ellipse ($0 < e_1 < 1$) and $e_2$ being the eccentricity of a hyperbola ($e_2 > 1$).
Since $e_1 e_2 = 2$, if $0 < e_1 < 1$, then $e_2 = 2/e_1 > 2$. This automatically satisfies $e_2 > 1$.
We need the roots of $t^2 - at + 2 = 0$ to satisfy $0 < e_1 < 1 < e_2$. This occurs when $f(1)$ has the opposite sign of the leading coefficient (which is positive) and the roots are real.
Conditions:
- Discriminant $D > 0 \implies a^2 > 8$.
- $f(1) < 0$. $f(1) = 3 - a$. So, $3 - a < 0 \implies a > 3$.
Combining $a^2 > 8$ and $a > 3$: The intersection is $a > 3$.
Therefore, the set is $(3, \infty)$. So, $\gamma = 3$.
Calculating $\alpha^2 + \beta^2 + \gamma^2$
We have $\alpha = 2\sqrt{2}$, $\beta = 3$, and $\gamma = 3$.
- $\alpha^2 = (2\sqrt{2})^2 = 4 \times 2 = 8$.
- $\beta^2 = 3^2 = 9$.
- $\gamma^2 = 3^2 = 9$.
The required sum is $\alpha^2 + \beta^2 + \gamma^2 = 8 + 9 + 9 = 26$.
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Important Questions from Algebra
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