2次方程式 $2x^2 - 4x + 1 = 0$ の2つの解を $\alpha$, $\beta$ とするとき、以下の式の値を求めます。 (1) $\alpha^2 + \alpha\beta + \beta^2$ (2) $\frac{\beta}{\alpha - 2} + \frac{\alpha}{\beta - 2}$ (3) $\alpha^4 + \beta^4$

1. 問題の内容

2次方程式 $2x^2 - 4x + 1 = 0$ の2つの解を $\alpha$, $\beta$ とするとき、以下の式の値を求めます。

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

(2) $\frac{\beta}{\alpha - 2} + \frac{\alpha}{\beta - 2}$

(3) $\alpha^4 + \beta^4$

2. 解き方の手順

まず、解と係数の関係から、

$\alpha + \beta = -\frac{-4}{2} = 2$

$\alpha \beta = \frac{1}{2}$

が得られます。

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

$= (2)^2 - \frac{1}{2} = 4 - \frac{1}{2} = \frac{7}{2}$

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

$= \frac{\beta^2 - 2\beta + \alpha^2 - 2\alpha}{\alpha\beta - 2(\alpha + \beta) + 4}$

$= \frac{\alpha^2 + \beta^2 - 2(\alpha + \beta)}{\alpha\beta - 2(\alpha + \beta) + 4}$

$= \frac{(\alpha + \beta)^2 - 2\alpha\beta - 2(\alpha + \beta)}{\alpha\beta - 2(\alpha + \beta) + 4}$

$= \frac{(2)^2 - 2(\frac{1}{2}) - 2(2)}{\frac{1}{2} - 2(2) + 4}$

$= \frac{4 - 1 - 4}{\frac{1}{2} - 4 + 4} = \frac{-1}{\frac{1}{2}} = -2$

(3) $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2$

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (2)^2 - 2(\frac{1}{2}) = 4 - 1 = 3$

$\alpha^4 + \beta^4 = (3)^2 - 2(\frac{1}{2})^2 = 9 - 2(\frac{1}{4}) = 9 - \frac{1}{2} = \frac{17}{2}$

3. 最終的な答え

(1) $\frac{7}{2}$

(2) $-2$

(3) $\frac{17}{2}$