与えられた8つの二次方程式を解く問題です。 (1) $4x^2 = 1$ (2) $(x-3)^2 - 5 = 0$ (3) $x^2 + 6x + 8 = 0$ (4) $x^2 + 6x + 9 = 0$ (5) $9x^2 - 12x + 4 = 0$ (6) $3x^2 - 4x - 4 = 0$ (7) $x^2 - 7x + 2 = 0$ (8) $2x^2 + 6x - 1 = 0$

1. 問題の内容

与えられた8つの二次方程式を解く問題です。

(1) $4x^2 = 1$

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

(3) $x^2 + 6x + 8 = 0$

(4) $x^2 + 6x + 9 = 0$

(5) $9x^2 - 12x + 4 = 0$

(6) $3x^2 - 4x - 4 = 0$

(7) $x^2 - 7x + 2 = 0$

(8) $2x^2 + 6x - 1 = 0$

2. 解き方の手順

(1) $4x^2 = 1$

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

$x = \pm \sqrt{\frac{1}{4}}$

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

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

$(x-3)^2 = 5$

$x-3 = \pm \sqrt{5}$

$x = 3 \pm \sqrt{5}$

(3) $x^2 + 6x + 8 = 0$

$(x+2)(x+4) = 0$

$x = -2, -4$

(4) $x^2 + 6x + 9 = 0$

$(x+3)^2 = 0$

$x = -3$

(5) $9x^2 - 12x + 4 = 0$

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

$3x-2 = 0$

$x = \frac{2}{3}$

(6) $3x^2 - 4x - 4 = 0$

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

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

(7) $x^2 - 7x + 2 = 0$

解の公式：$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(2)}}{2(1)}$

$x = \frac{7 \pm \sqrt{49 - 8}}{2}$

$x = \frac{7 \pm \sqrt{41}}{2}$

(8) $2x^2 + 6x - 1 = 0$

解の公式：$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-6 \pm \sqrt{6^2 - 4(2)(-1)}}{2(2)}$

$x = \frac{-6 \pm \sqrt{36 + 8}}{4}$

$x = \frac{-6 \pm \sqrt{44}}{4}$

$x = \frac{-6 \pm 2\sqrt{11}}{4}$

$x = \frac{-3 \pm \sqrt{11}}{2}$

3. 最終的な答え

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

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

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

(4) $x = -3$

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

(6) $x = -\frac{2}{3}, 2$

(7) $x = \frac{7 \pm \sqrt{41}}{2}$

(8) $x = \frac{-3 \pm \sqrt{11}}{2}$