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:
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:
- \( 10\sqrt{2}P_9 \)
- \( 10\sqrt{3}P_9 \)
- \( 11\sqrt{2}P_9 \)
- \( 11\sqrt{3}P_9 \)
The Correct Option is B
Approach Solution - 1
To solve this problem, we need to evaluate the expression:
Given that \( \alpha, \beta \) are roots of the quadratic equation:
\[x^2 - \sqrt{2} x - \sqrt{3} = 0\]Using Vieta's formulas, we have:
- Sum of the roots: \(\alpha + \beta = \sqrt{2}\)
- Product of the roots: \(\alpha \beta = -\sqrt{3}\)
The sequence \(P_n = \alpha^n - \beta^n\) satisfies the recurrence relation:
\[P_{n} = (\alpha + \beta) P_{n-1} - \alpha \beta P_{n-2}\]Substitute the known values:
\[P_n = \sqrt{2} P_{n-1} + \sqrt{3} P_{n-2}\]This recurrence allows us to express any term in the sequence in terms of the previous two terms:
\[P_{10} = \sqrt{2} P_9 + \sqrt{3} P_8, \quad P_{11} = \sqrt{2} P_{10} + \sqrt{3} P_9, \quad P_{12} = \sqrt{2} P_{11} + \sqrt{3} P_{10}\]We aim to evaluate:
\[\left( 11\sqrt{3} - 10\sqrt{2} \right) P_{10} + \left( 11\sqrt{2} + 10 \right) P_{11} - 11 P_{12}\]Substitute the recurrence relations for \( P_{10} \), \( P_{11} \), and \( P_{12} \):
- \(P_{10} = \sqrt{2} P_9 + \sqrt{3} P_8\)
- \(P_{11} = \sqrt{2} (\sqrt{2} P_9 + \sqrt{3} P_8) + \sqrt{3} P_9 = 2 P_9 + \sqrt{6} P_8\)
- \(P_{12} = \sqrt{2} \cdot (2 P_9 + \sqrt{6} P_8) + \sqrt{3} \cdot (\sqrt{2} P_9 + \sqrt{3} P_8)\)
Simplifying these, obtain the combinations back to \( P_9 \) and \( P_8 \).
Now substitute into the expression:
\[\left( 11\sqrt{3} - 10\sqrt{2} \right) (\sqrt{2} P_9 + \sqrt{3} P_8) + \left( 11\sqrt{2} + 10 \right) (2 P_9 + \sqrt{6} P_8) - 11 P_{12}\]Group terms to factor out \(P_9\) and \(P_8\), then simplify to the desired result:
After computations, it reduces to:
\[10\sqrt{3} P_9\]Thus, the expression is equal to the option \(10\sqrt{3}P_9\).
Approach Solution -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:
\[ 10\sqrt{3}P_9 \]
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