2次方程式 $2x^2 + 3x - 4 = 0$ の2つの解を $\alpha$, $\beta$ とするとき、以下の式の値を求めよ。 (1) $(\alpha - 2)(\beta - 2)$ (2) $\alpha^2\beta + \alpha\beta^2$ (3) $\frac{1}{\alpha} + \frac{1}{\beta}$ (4) $\alpha^2 + \beta^2$ (5) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta}$ (6) $\alpha^3 + \beta^3$

1. 問題の内容

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

(1) $(\alpha - 2)(\beta - 2)$

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

(3) $\frac{1}{\alpha} + \frac{1}{\beta}$

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

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

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

2. 解き方の手順

解と係数の関係より、

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

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

である。

(1) $(\alpha - 2)(\beta - 2) = \alpha\beta - 2(\alpha + \beta) + 4 = -2 - 2(-\frac{3}{2}) + 4 = -2 + 3 + 4 = 5$

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

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

(4) $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-\frac{3}{2})^2 - 2(-2) = \frac{9}{4} + 4 = \frac{9 + 16}{4} = \frac{25}{4}$

(5) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\beta^2 + \alpha^2}{\alpha\beta} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{\frac{25}{4}}{-2} = -\frac{25}{8}$

(6) $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) = (-\frac{3}{2})((-\frac{3}{2})^2 - 3(-2)) = (-\frac{3}{2})(\frac{9}{4} + 6) = (-\frac{3}{2})(\frac{9 + 24}{4}) = (-\frac{3}{2})(\frac{33}{4}) = -\frac{99}{8}$

3. 最終的な答え

(1) 5

(2) 3

(3) $\frac{3}{4}$

(4) $\frac{25}{4}$

(5) $-\frac{25}{8}$

(6) $-\frac{99}{8}$