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 $\alpha^2 + \beta^2 + \gamma^2 = 8 + 9 + 9 = 26$.
Similar Questions
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
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Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
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Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
