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

1. 問題の内容

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

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

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

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

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

2. 解き方の手順

まず、解と係数の関係から、$\alpha + \beta$ と $\alpha\beta$ の値を求める。

$2x^2 - 6x - 3 = 0$ において、解と係数の関係より、

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

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

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

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

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 3^2 - 2(-\frac{3}{2}) = 9 + 3 = 12$

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

$(\alpha - \beta)^2 = \alpha^2 - 2\alpha\beta + \beta^2 = \alpha^2 + \beta^2 - 2\alpha\beta = 12 - 2(-\frac{3}{2}) = 12 + 3 = 15$

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

$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) = 3(3^2 - 3(-\frac{3}{2})) = 3(9 + \frac{9}{2}) = 3(\frac{18+9}{2}) = 3(\frac{27}{2}) = \frac{81}{2}$

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

$\frac{\beta^2}{\alpha} + \frac{\alpha^2}{\beta} = \frac{\beta^3 + \alpha^3}{\alpha\beta} = \frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{\frac{81}{2}}{-\frac{3}{2}} = \frac{81}{2} \times (-\frac{2}{3}) = -27$

3. 最終的な答え

(1) $\alpha^2 + \beta^2 = 12$

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

(3) $\alpha^3 + \beta^3 = \frac{81}{2}$

(4) $\frac{\beta^2}{\alpha} + \frac{\alpha^2}{\beta} = -27$