2次方程式 $x^2 + 5x + 7 = 0$ の2つの解を $\alpha, \beta$ とするとき、次の式の値を求めよ。 (1) $\alpha^2 \beta + \alpha \beta^2$ (2) $\frac{1}{\alpha} + \frac{1}{\beta}$ (3) $(\alpha + 1)(\beta + 1)$ (4) $\alpha^2 + \beta^2$ (5) $(\alpha - \beta)^2$ (6) $\alpha^3 + \beta^3$

1. 問題の内容

2次方程式 $x^2 + 5x + 7 = 0$ の2つの解を $\alpha, \beta$ とするとき、次の式の値を求めよ。

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

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

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

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

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

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

2. 解き方の手順

まず、解と係数の関係より、

$\alpha + \beta = -5$

$\alpha \beta = 7$

(1) $\alpha^2 \beta + \alpha \beta^2 = \alpha \beta (\alpha + \beta) = 7(-5) = -35$

(2) $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-5}{7} = -\frac{5}{7}$

(3) $(\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1 = 7 + (-5) + 1 = 3$

(4) $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = (-5)^2 - 2(7) = 25 - 14 = 11$

(5) $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = (-5)^2 - 4(7) = 25 - 28 = -3$

(6) $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta) = (-5)^3 - 3(7)(-5) = -125 + 105 = -20$

3. 最終的な答え

(1) -35

(2) -5/7

(3) 3

(4) 11

(5) -3

(6) -20