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

1. 問題の内容

2次方程式 $x^2 + 3x + 5 = 0$ の2つの解を $\alpha$, $\beta$ とするとき、次の式の値を求める。

(1) $\alpha + \beta$

(2) $\alpha \beta$

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

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

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

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

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

2. 解き方の手順

解と係数の関係より、

$\alpha + \beta = -\frac{3}{1} = -3$

$\alpha \beta = \frac{5}{1} = 5$

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

(2) $\alpha \beta = 5$

(3) $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = (-3)^2 - 2(5) = 9 - 10 = -1$

(4) $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta) = (-3)((-3)^2 - 3(5)) = (-3)(9 - 15) = (-3)(-6) = 18$

(5) $\alpha \beta^2 + \alpha^2 \beta = \alpha \beta (\beta + \alpha) = \alpha \beta (\alpha + \beta) = 5(-3) = -15$

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

(7) $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = (-3)^2 - 4(5) = 9 - 20 = -11$

3. 最終的な答え

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

(2) $\alpha \beta = 5$

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

(4) $\alpha^3 + \beta^3 = 18$

(5) $\alpha \beta^2 + \alpha^2 \beta = -15$

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

(7) $(\alpha - \beta)^2 = -11$