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

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、次の値を求めよ。

(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

解と係数の関係より、

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

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

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

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

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

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

3. 最終的な答え

(1) 6

(2) $-\frac{23}{4}$

(3) $-\frac{23}{16}$

(4) $-\frac{117}{8}$