Question

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

Answer 1

4

Answer 2

Let $\alpha, \beta$ be the roots of the quadratic equation $x^2 + \sqrt{2}x - 8 = 0$. Since $\alpha$ and $\beta$ are roots of the equation, they must satisfy the equation. Thus, we have: $\alpha^2 + \sqrt{2}\alpha - 8 = 0 \implies \alpha^2 + \sqrt{2}\alpha = 8 \quad (1)$ $\beta^2 + \sqrt{2}\beta - 8 = 0 \implies \beta^2 + \sqrt{2}\beta = 8 \quad (2)$

We are given $U_n = \alpha^n + \beta^n$. We need to find the value of $\frac{U_{10} + \sqrt{2}U_9}{2U_8}$.

Substitute the definition of $U_n$ into the expression: $\frac{U_{10} + \sqrt{2}U_9}{2U_8} = \frac{(\alpha^{10} + \beta^{10}) + \sqrt{2}(\alpha^9 + \beta^9)}{2(\alpha^8 + \beta^8)}$

Expand the numerator: $= \frac{\alpha^{10} + \beta^{10} + \sqrt{2}\alpha^9 + \sqrt{2}\beta^9}{2(\alpha^8 + \beta^8)}$

Rearrange the terms in the numerator by grouping terms involving $\alpha$ and $\beta$: $= \frac{(\alpha^{10} + \sqrt{2}\alpha^9) + (\beta^{10} + \sqrt{2}\beta^9)}{2(\alpha^8 + \beta^8)}$

Factor out $\alpha^8$ from the first parenthesis and $\beta^8$ from the second parenthesis in the numerator: $= \frac{\alpha^8(\alpha^2 + \sqrt{2}\alpha) + \beta^8(\beta^2 + \sqrt{2}\beta)}{2(\alpha^8 + \beta^8)}$

Now, substitute the relations from (1) and (2) into this expression: $= \frac{\alpha^8(8) + \beta^8(8)}{2(\alpha^8 + \beta^8)}$

Factor out 8 from the numerator: $= \frac{8(\alpha^8 + \beta^8)}{2(\alpha^8 + \beta^8)}$

Since the roots $\alpha, \beta$ are real and non-zero (as the constant term is -8), $\alpha^8 + \beta^8$ will not be zero. Thus, we can cancel the term $(\alpha^8 + \beta^8)$ from the numerator and denominator: $= \frac{8}{2}$ $= 4$