与えられた6つの二次方程式を解く問題です。 (1) $x^2 = 24$ (2) $\frac{1}{10}x^2 - 10 = 0$ (3) $4x^2 - 13 = -11$ (4) $(x-5)^2 = 2$ (5) $(x-4)^2 - 18 = 0$ (6) $(2x-5)^2 = 36$

1. 問題の内容

与えられた6つの二次方程式を解く問題です。

(1) $x^2 = 24$

(2) $\frac{1}{10}x^2 - 10 = 0$

(3) $4x^2 - 13 = -11$

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

(5) $(x-4)^2 - 18 = 0$

(6) $(2x-5)^2 = 36$

2. 解き方の手順

(1) $x^2 = 24$

両辺の平方根をとると、

$x = \pm \sqrt{24} = \pm \sqrt{4 \cdot 6} = \pm 2\sqrt{6}$

(2) $\frac{1}{10}x^2 - 10 = 0$

$\frac{1}{10}x^2 = 10$

$x^2 = 100$

$x = \pm \sqrt{100} = \pm 10$

(3) $4x^2 - 13 = -11$

$4x^2 = 2$

$x^2 = \frac{2}{4} = \frac{1}{2}$

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

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

$x-5 = \pm \sqrt{2}$

$x = 5 \pm \sqrt{2}$

(5) $(x-4)^2 - 18 = 0$

$(x-4)^2 = 18$

$x-4 = \pm \sqrt{18} = \pm \sqrt{9 \cdot 2} = \pm 3\sqrt{2}$

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

(6) $(2x-5)^2 = 36$

$2x-5 = \pm \sqrt{36} = \pm 6$

$2x = 5 \pm 6$

$x = \frac{5 \pm 6}{2}$

$x = \frac{5+6}{2} = \frac{11}{2}$ または $x = \frac{5-6}{2} = \frac{-1}{2} = -\frac{1}{2}$

3. 最終的な答え

(1) $x = \pm 2\sqrt{6}$

(2) $x = \pm 10$

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

(4) $x = 5 \pm \sqrt{2}$

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

(6) $x = \frac{11}{2}, -\frac{1}{2}$