与えられた6つの方程式をそれぞれ解き、$x$ の値を求める。 (1) $x^2 - 15 = 0$ (2) $5x^2 - 60 = 0$ (3) $3x^2 - 1 = 35$ (4) $(x+6)^2 = 48$ (5) $16(x-3)^2 = 3$ (6) $25(x+2)^2 = 64$

1. 問題の内容

与えられた6つの方程式をそれぞれ解き、$x$ の値を求める。

(1) $x^2 - 15 = 0$

(2) $5x^2 - 60 = 0$

(3) $3x^2 - 1 = 35$

(4) $(x+6)^2 = 48$

(5) $16(x-3)^2 = 3$

(6) $25(x+2)^2 = 64$

2. 解き方の手順

(1) $x^2 - 15 = 0$

$x^2 = 15$

$x = \pm\sqrt{15}$

(2) $5x^2 - 60 = 0$

$5x^2 = 60$

$x^2 = 12$

$x = \pm\sqrt{12} = \pm2\sqrt{3}$

(3) $3x^2 - 1 = 35$

$3x^2 = 36$

$x^2 = 12$

$x = \pm\sqrt{12} = \pm2\sqrt{3}$

(4) $(x+6)^2 = 48$

$x+6 = \pm\sqrt{48} = \pm4\sqrt{3}$

$x = -6 \pm 4\sqrt{3}$

(5) $16(x-3)^2 = 3$

$(x-3)^2 = \frac{3}{16}$

$x-3 = \pm\sqrt{\frac{3}{16}} = \pm\frac{\sqrt{3}}{4}$

$x = 3 \pm \frac{\sqrt{3}}{4} = \frac{12 \pm \sqrt{3}}{4}$

(6) $25(x+2)^2 = 64$

$(x+2)^2 = \frac{64}{25}$

$x+2 = \pm\sqrt{\frac{64}{25}} = \pm\frac{8}{5}$

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

$x = \frac{-10+8}{5} = \frac{-2}{5}$ または $x = \frac{-10-8}{5} = \frac{-18}{5}$

3. 最終的な答え

(1) $x = \pm\sqrt{15}$

(2) $x = \pm2\sqrt{3}$

(3) $x = \pm2\sqrt{3}$

(4) $x = -6 \pm 4\sqrt{3}$

(5) $x = \frac{12 \pm \sqrt{3}}{4}$

(6) $x = -\frac{2}{5}, -\frac{18}{5}$