2次方程式 $2x^2 - 2x + 3 = 0$ の2つの解を $\alpha, \beta$ とするとき、以下の値を求めます。 (1) $\alpha + \beta$ (2) $\alpha \beta$ (3) $\alpha^2 + \beta^2$ (4) $2\alpha^2\beta + 2\alpha\beta^2$ (5) $4(3-\alpha)(3-\beta)$ (6) $2(\alpha^3 + \beta^3)$ (7) $\frac{1}{\alpha} + \frac{1}{\beta}$ (8) $9(\frac{1}{\alpha^2} + \frac{1}{\beta^2})$

1. 問題の内容

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

(1) $\alpha + \beta$

(2) $\alpha \beta$

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

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

(5) $4(3-\alpha)(3-\beta)$

(6) $2(\alpha^3 + \beta^3)$

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

(8) $9(\frac{1}{\alpha^2} + \frac{1}{\beta^2})$

2. 解き方の手順

まず、解と係数の関係より、

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

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

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

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

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

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

(5) $4(3-\alpha)(3-\beta) = 4(9 - 3\alpha - 3\beta + \alpha\beta) = 4(9 - 3(\alpha+\beta) + \alpha\beta) = 4(9 - 3(1) + \frac{3}{2}) = 4(6 + \frac{3}{2}) = 4(\frac{15}{2}) = 30$

(6) $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) = (1)(1^2 - 3(\frac{3}{2})) = 1 - \frac{9}{2} = -\frac{7}{2}$

$2(\alpha^3 + \beta^3) = 2(-\frac{7}{2}) = -7$

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

(8) $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{-2}{(\frac{3}{2})^2} = \frac{-2}{\frac{9}{4}} = -2 \cdot \frac{4}{9} = -\frac{8}{9}$

$9(\frac{1}{\alpha^2} + \frac{1}{\beta^2}) = 9(-\frac{8}{9}) = -8$

3. 最終的な答え

(1) 1

(2) $\frac{3}{2}$

(3) -2

(4) 3

(5) 30

(6) -7

(7) $\frac{2}{3}$

(8) -8