2次方程式 $2x^2 - 3x + 8 = 0$ の2つの解を $\alpha, \beta$ とするとき、以下の式の値を求めよ。 (1) $\alpha^2\beta + \alpha\beta^2$ (2) $\alpha^2 + \beta^2$ (3) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta}$ (4) $\alpha^3 + \beta^3$

## 数学の問題の回答

1. 問題の内容

2次方程式 $2x^2 - 3x + 8 = 0$ の2つの解を $\alpha, \beta$ とするとき、以下の式の値を求めよ。

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

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

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

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

2. 解き方の手順

まず、解と係数の関係より、以下の関係が成り立つ。

$\alpha + \beta = -\frac{-3}{2} = \frac{3}{2}$

$\alpha\beta = \frac{8}{2} = 4$

(1) $\alpha^2\beta + \alpha\beta^2 = \alpha\beta(\alpha + \beta) = 4 \cdot \frac{3}{2} = 6$

(2) $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (\frac{3}{2})^2 - 2 \cdot 4 = \frac{9}{4} - 8 = \frac{9}{4} - \frac{32}{4} = -\frac{23}{4}$

(3) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{-\frac{23}{4}}{4} = -\frac{23}{16}$

(4) $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) = \frac{3}{2} ((\frac{3}{2})^2 - 3 \cdot 4) = \frac{3}{2}(\frac{9}{4} - 12) = \frac{3}{2}(\frac{9}{4} - \frac{48}{4}) = \frac{3}{2}(-\frac{39}{4}) = -\frac{117}{8}$

3. 最終的な答え

(1) $\alpha^2\beta + \alpha\beta^2 = 6$

(2) $\alpha^2 + \beta^2 = -\frac{23}{4}$

(3) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = -\frac{23}{16}$

(4) $\alpha^3 + \beta^3 = -\frac{117}{8}$