次の方程式を解く問題です。 (1) $x^2 - 18 = 0$ (2) $2x^2 - 32 = 0$ (3) $5x^2 - 8 = 0$ (4) $(x - 3)^2 - 12 = 0$ (5) $(x - 1)^2 - 4 = 0$ (6) $5(x + 3)^2 - 45 = 0$

1. 問題の内容

次の方程式を解く問題です。

(1) $x^2 - 18 = 0$

(2) $2x^2 - 32 = 0$

(3) $5x^2 - 8 = 0$

(4) $(x - 3)^2 - 12 = 0$

(5) $(x - 1)^2 - 4 = 0$

(6) $5(x + 3)^2 - 45 = 0$

2. 解き方の手順

(1) $x^2 - 18 = 0$

$x^2 = 18$

$x = \pm \sqrt{18} = \pm \sqrt{9 \times 2} = \pm 3\sqrt{2}$

(2) $2x^2 - 32 = 0$

$2x^2 = 32$

$x^2 = 16$

$x = \pm \sqrt{16} = \pm 4$

(3) $5x^2 - 8 = 0$

$5x^2 = 8$

$x^2 = \frac{8}{5}$

$x = \pm \sqrt{\frac{8}{5}} = \pm \frac{\sqrt{8}}{\sqrt{5}} = \pm \frac{2\sqrt{2}}{\sqrt{5}} = \pm \frac{2\sqrt{10}}{5}$

(4) $(x - 3)^2 - 12 = 0$

$(x - 3)^2 = 12$

$x - 3 = \pm \sqrt{12} = \pm 2\sqrt{3}$

$x = 3 \pm 2\sqrt{3}$

(5) $(x - 1)^2 - 4 = 0$

$(x - 1)^2 = 4$

$x - 1 = \pm \sqrt{4} = \pm 2$

$x = 1 \pm 2$

$x = 1 + 2 = 3$ or $x = 1 - 2 = -1$

(6) $5(x + 3)^2 - 45 = 0$

$5(x + 3)^2 = 45$

$(x + 3)^2 = 9$

$x + 3 = \pm \sqrt{9} = \pm 3$

$x = -3 \pm 3$

$x = -3 + 3 = 0$ or $x = -3 - 3 = -6$

3. 最終的な答え

(1) $x = \pm 3\sqrt{2}$

(2) $x = \pm 4$

(3) $x = \pm \frac{2\sqrt{10}}{5}$

(4) $x = 3 \pm 2\sqrt{3}$

(5) $x = 3, -1$

(6) $x = 0, -6$