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

1. 問題の内容

2次方程式 $x^2 + 6x - 3 = 0$ の2つの解を $\alpha$, $\beta$ とするとき、以下の式の値を求めます。

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

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

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

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

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

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

2. 解き方の手順

まず、解と係数の関係から、

$\alpha + \beta = -6$

$\alpha\beta = -3$

がわかります。これらを利用して各式の値を求めます。

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

$(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$ より、

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

$\alpha^2 + \beta^2 = (-6)^2 - 2(-3) = 36 + 6 = 42$

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

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

$\alpha^2\beta + \alpha\beta^2 = (-3)(-6) = 18$

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

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

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

$\alpha^3 + \beta^3 = (-6)((-6)^2 - 3(-3)) = (-6)(36 + 9) = (-6)(45) = -270$

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

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

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{-6}{-3} = 2$

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

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

$(\alpha - \beta)^2 = (-6)^2 - 4(-3) = 36 + 12 = 48$

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

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

$(\alpha + 1)(\beta + 1) = -3 + (-6) + 1 = -8$

3. 最終的な答え

(1) $\alpha^2 + \beta^2 = 42$

(2) $\alpha^2\beta + \alpha\beta^2 = 18$

(3) $\alpha^3 + \beta^3 = -270$

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

(5) $(\alpha - \beta)^2 = 48$

(6) $(\alpha + 1)(\beta + 1) = -8$