Question

Let $\alpha, \beta$ be roots of $x^2 + \sqrt{2}x - 8 = 0$. If $U_n = \alpha^n + \beta^n$, then \[ \frac{U_{10} + \sqrt{12} U_9}{2 U_8} \] is equal to ________.

Answer 1

Correct Answer: 4

The problem provides a quadratic equation \(x^2 + \sqrt{2}x - 8 = 0\) with roots \(\alpha\) and \(\beta\). A sequence \(U_n\) is defined as \(U_n = \alpha^n + \beta^n\). We are asked to find the value of a given expression involving terms of this sequence. It is a common pattern in such problems that the expression is designed to simplify using a recurrence relation derived from the quadratic equation. The given expression \(\frac{U_{10} + \sqrt{12} U_9}{2 U_8}\) appears to contain a typo, as \(\sqrt{12}\) does not align with the coefficient \(\sqrt{2}\) in the quadratic equation. The intended expression for a clean solution is almost certainly \(\frac{U_{10} + \sqrt{2} U_9}{2 U_8}\). We will proceed by solving this corrected version.

Concept Used:

If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\), they must satisfy the equation. This property can be used to establish a linear recurrence relation for the sequence \(U_n = \alpha^n + \beta^n\). Specifically, multiplying the equation \(a\alpha^2 + b\alpha + c = 0\) by \(\alpha^{n-2}\) and \(a\beta^2 + b\beta + c = 0\) by \(\beta^{n-2}\) and then adding them yields the relation \(aU_n + bU_{n-1} + cU_{n-2} = 0\).

Step-by-Step Solution:

Step 1: Establish the equations satisfied by the roots \(\alpha\) and \(\beta\).

Since \(\alpha\) and \(\beta\) are the roots of \(x^2 + \sqrt{2}x - 8 = 0\), they satisfy the equation:

\[ \alpha^2 + \sqrt{2}\alpha - 8 = 0 \] \[ \beta^2 + \sqrt{2}\beta - 8 = 0 \]

Step 2: Derive the recurrence relation for the sequence \(U_n\).

Multiply the first equation by \(\alpha^{n-2}\) and the second equation by \(\beta^{n-2}\) (for \(n \ge 2\)):

\[ \alpha^n + \sqrt{2}\alpha^{n-1} - 8\alpha^{n-2} = 0 \] \[ \beta^n + \sqrt{2}\beta^{n-1} - 8\beta^{n-2} = 0 \]

Adding these two equations together:

\[ (\alpha^n + \beta^n) + \sqrt{2}(\alpha^{n-1} + \beta^{n-1}) - 8(\alpha^{n-2} + \beta^{n-2}) = 0 \]

Using the definition \(U_n = \alpha^n + \beta^n\), we get the recurrence relation:

\[ U_n + \sqrt{2}U_{n-1} - 8U_{n-2} = 0 \]

Step 3: Apply the recurrence relation for \(n=10\).

Substitute \(n = 10\) into the recurrence relation:

\[ U_{10} + \sqrt{2}U_9 - 8U_8 = 0 \]

Rearranging this equation gives a direct relationship between the terms in the numerator of the expression we want to evaluate:

\[ U_{10} + \sqrt{2}U_9 = 8U_8 \]

Step 4: Substitute the result from the recurrence into the expression.

The expression to be evaluated (with the likely correction) is:

\[ \frac{U_{10} + \sqrt{2} U_9}{2 U_8} \]

Substitute the expression for the numerator that we found in Step 3:

\[ \frac{8U_8}{2U_8} \]

Step 5: Calculate the final value.

Assuming \(U_8 \neq 0\), we can cancel the term from the numerator and the denominator:

\[ \frac{8}{2} = 4 \]

The value of the expression is 4.

Answer 2

We have:
\[U_{10} = \alpha^{10} + \beta^{10} \quad \text{and} \quad U_9 = \alpha^9 + \beta^9\]
Then:
\[\frac{U_{10} + \sqrt{12} U_9}{2 U_8} = \frac{\alpha^{10} + \beta^{10} + \sqrt{2} (\alpha^9 + \beta^9)}{2 \left( \alpha^8 + \beta^8 \right)}\]
Further simplifying, we get:
\[= \frac{\alpha^8 \left( \alpha^2 + \sqrt{2} \alpha \right) + \beta^8 \left( \beta^2 + \sqrt{2} \beta \right)}{2 \left( \alpha^8 + \beta^8 \right)}\]
Grouping terms, we have:
\[= \frac{8 \alpha^8 + 8 \beta^8}{2 \left( \alpha^8 + \beta^8 \right)} = 4\]