次の方程式を解く問題です。 (1) $8x^3 - 1 = 0$ (2) $2x^4 + x^2 - 6 = 0$ (3) $x(x+1)(x+2) = 3 \cdot 4 \cdot 5$ (4) $(x^2 - x)^2 - 8(x^2 - x) + 12 = 0$

1. 問題の内容

次の方程式を解く問題です。

(1) $8x^3 - 1 = 0$

(2) $2x^4 + x^2 - 6 = 0$

(3) $x(x+1)(x+2) = 3 \cdot 4 \cdot 5$

(4) $(x^2 - x)^2 - 8(x^2 - x) + 12 = 0$

2. 解き方の手順

(1) $8x^3 - 1 = 0$

$8x^3 = 1$

$x^3 = \frac{1}{8}$

$x = \sqrt[3]{\frac{1}{8}}$

$x = \frac{1}{2}$

(2) $2x^4 + x^2 - 6 = 0$

$x^2 = t$とおくと

$2t^2 + t - 6 = 0$

$(2t - 3)(t + 2) = 0$

$t = \frac{3}{2}, -2$

$x^2 = \frac{3}{2}, -2$

$x = \pm \sqrt{\frac{3}{2}}, \pm \sqrt{-2}$

$x = \pm \frac{\sqrt{6}}{2}, \pm i\sqrt{2}$

(3) $x(x+1)(x+2) = 3 \cdot 4 \cdot 5$

$x(x+1)(x+2) = 60$

$x(x^2 + 3x + 2) = 60$

$x^3 + 3x^2 + 2x - 60 = 0$

$(x - 3)(x^2 + 6x + 20) = 0$

$x = 3$または$x^2 + 6x + 20 = 0$

$x = \frac{-6 \pm \sqrt{36 - 80}}{2}$

$x = \frac{-6 \pm \sqrt{-44}}{2}$

$x = \frac{-6 \pm 2i\sqrt{11}}{2}$

$x = -3 \pm i\sqrt{11}$

$x = 3, -3 \pm i\sqrt{11}$

(4) $(x^2 - x)^2 - 8(x^2 - x) + 12 = 0$

$x^2 - x = t$とおくと

$t^2 - 8t + 12 = 0$

$(t - 2)(t - 6) = 0$

$t = 2, 6$

$x^2 - x = 2, x^2 - x = 6$

$x^2 - x - 2 = 0, x^2 - x - 6 = 0$

$(x - 2)(x + 1) = 0, (x - 3)(x + 2) = 0$

$x = 2, -1, 3, -2$

3. 最終的な答え

(1) $x = \frac{1}{2}$

(2) $x = \pm \frac{\sqrt{6}}{2}, \pm i\sqrt{2}$

(3) $x = 3, -3 \pm i\sqrt{11}$

(4) $x = 2, -1, 3, -2$