Question

Let $\alpha, \beta; \, \alpha > \beta$, be the roots of the equation $x^2 - \sqrt{2}x - \sqrt{3} = 0.$ Let $P_n = \alpha^n - \beta^n, \, n \in \mathbb{N}$. Then $\left( 11\sqrt{3} - 10\sqrt{2} \right) P_{10} + \left( 11\sqrt{2} + 10 \right) P_{11} - 11P_{12}$ is equal to:

• A. $10\sqrt{2}P_9$
• B. $10\sqrt{3}P_9$
• C. $11\sqrt{2}P_9$
• D. $11\sqrt{3}P_9$

Answer 1

Answer: (B)

$10\sqrt{3}P_9$

Answer 2

We are given that α and β are the roots of the quadratic equation:

$x^2 - \sqrt{2}x - \sqrt{3} = 0$

Step 1: Find the roots α and β
To find α and β, we use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the given equation, $a = 1$, $b = -\sqrt{2}$, and $c = -\sqrt{3}$. Substituting these values into the quadratic formula:

$x = \frac{\sqrt{2} \pm \sqrt{(\sqrt{2})^2 - 4(1)(-\sqrt{3})}}{2(1)}$

$x = \frac{\sqrt{2} \pm \sqrt{2 + 4\sqrt{3}}}{2}$

Thus, the roots α and β are:

$\alpha = \frac{\sqrt{2} + \sqrt{2 + 4\sqrt{3}}}{2}, \quad \beta = \frac{\sqrt{2} - \sqrt{2 + 4\sqrt{3}}}{2}$

Step 2: Use recurrence relation for $P_n$
We are given that $P_n = \alpha^n - \beta^n$. From the given quadratic equation, we know that:

$\alpha + \beta = \sqrt{2}, \quad \alpha\beta = -\sqrt{3}$

Using this, we can derive a recurrence relation for $P_n$. The recurrence relation is:

$P_n = (\alpha + \beta)P_{n-1} - \alpha\beta P_{n-2}$

Substituting the values $\alpha + \beta = \sqrt{2}$ and $\alpha\beta = -\sqrt{3}$, we get:

$P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2}$

Step 3: Calculate the required expression
Now, we need to calculate the following expression:

$(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$

Using the recurrence relation for $P_n$, we can express each term in terms of $P_9$:

$P_{10} = \sqrt{2}P_9 + \sqrt{3}P_8$

$P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9$

$P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$

Substituting these into the original expression, and simplifying, we find that the value of the expression is:

$10\sqrt{3}P_9$

Thus, the correct answer is:

$10\sqrt{3}P_9$