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

1. 問題の内容

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

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

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

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

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

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

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

2. 解き方の手順

解と係数の関係より、

$\alpha + \beta = 4$

$\alpha \beta = -2$

(1) $\alpha^2 \beta + \alpha \beta^2 = \alpha \beta (\alpha + \beta) = (-2)(4) = -8$

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

(3) $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2 \alpha \beta = (4)^2 - 2(-2) = 16 + 4 = 20$

(4) $(\alpha-\beta)^2 = (\alpha + \beta)^2 - 4 \alpha \beta = (4)^2 - 4(-2) = 16 + 8 = 24$

(5) $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2) = (\alpha + \beta)((\alpha+\beta)^2 - 3\alpha\beta) = (4)((4)^2 - 3(-2)) = 4(16 + 6) = 4(22) = 88$

(6) $\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{(\alpha + \beta)^2 - 2 \alpha \beta}{\alpha \beta} = \frac{(4)^2 - 2(-2)}{-2} = \frac{16+4}{-2} = \frac{20}{-2} = -10$

3. 最終的な答え

(1) -8

(2) 3

(3) 20

(4) 24

(5) 88

(6) -10