$\alpha = \frac{1}{\sqrt{5}-2}$、$\beta = \frac{1}{\sqrt{5}+2}$ のとき、以下の値をそれぞれ求めます。 (1) $\alpha^2 + \beta^2$ (2) $\alpha^3 + \beta^3$ (3) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta}$

1. 問題の内容

$\alpha = \frac{1}{\sqrt{5}-2}$、$\beta = \frac{1}{\sqrt{5}+2}$ のとき、以下の値をそれぞれ求めます。

(1) $\alpha^2 + \beta^2$

(2) $\alpha^3 + \beta^3$

(3) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta}$

2. 解き方の手順

まず、$\alpha$と$\beta$を簡単にします。

$\alpha = \frac{1}{\sqrt{5}-2} = \frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)} = \frac{\sqrt{5}+2}{5-4} = \sqrt{5}+2$

$\beta = \frac{1}{\sqrt{5}+2} = \frac{\sqrt{5}-2}{(\sqrt{5}+2)(\sqrt{5}-2)} = \frac{\sqrt{5}-2}{5-4} = \sqrt{5}-2$

(1) $\alpha^2 + \beta^2$ を計算します。

$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$

$\alpha + \beta = (\sqrt{5}+2) + (\sqrt{5}-2) = 2\sqrt{5}$

$\alpha\beta = (\sqrt{5}+2)(\sqrt{5}-2) = 5-4 = 1$

よって、

$\alpha^2 + \beta^2 = (2\sqrt{5})^2 - 2(1) = 4(5) - 2 = 20 - 2 = 18$

(2) $\alpha^3 + \beta^3$ を計算します。

$\alpha^3 + \beta^3 = (\alpha+\beta)(\alpha^2 - \alpha\beta + \beta^2) = (\alpha+\beta)((\alpha+\beta)^2 - 3\alpha\beta)$

$\alpha + \beta = 2\sqrt{5}$、$\alpha\beta = 1$ より、

$\alpha^3 + \beta^3 = (2\sqrt{5})((2\sqrt{5})^2 - 3(1)) = (2\sqrt{5})(20-3) = (2\sqrt{5})(17) = 34\sqrt{5}$

(3) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta}$ を計算します。

$\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^2 + \alpha^2}{\alpha\beta}$

$\alpha^2 + \beta^2 = 18$、$\alpha\beta = 1$ より、

$\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{18}{1} = 18$

3. 最終的な答え

(1) $\alpha^2 + \beta^2 = 18$

(2) $\alpha^3 + \beta^3 = 34\sqrt{5}$

(3) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = 18$