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

1. 問題の内容

与えられた6つの2次方程式を解く問題です。

(1) $3x^2 + 7x + 1 = 0$

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

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

(4) $2x^2 - 4x - 7 = 0$

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

(6) $x^2 - 2\sqrt{3}x + 3 = 0$

2. 解き方の手順

2次方程式 $ax^2 + bx + c = 0$ の解は、解の公式を用いて求めることができます。解の公式は次の通りです。

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

(1) $3x^2 + 7x + 1 = 0$

$a = 3$, $b = 7$, $c = 1$

$x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{-7 \pm \sqrt{49 - 12}}{6} = \frac{-7 \pm \sqrt{37}}{6}$

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

$a = 1$, $b = -3$, $c = -2$

$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$

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

$a = 1$, $b = 2$, $c = -1$

$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$

(4) $2x^2 - 4x - 7 = 0$

$a = 2$, $b = -4$, $c = -7$

$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2} = \frac{4 \pm \sqrt{16 + 56}}{4} = \frac{4 \pm \sqrt{72}}{4} = \frac{4 \pm 6\sqrt{2}}{4} = \frac{2 \pm 3\sqrt{2}}{2}$

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

$a = 9$, $b = -12$, $c = 4$

$x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 9 \cdot 4}}{2 \cdot 9} = \frac{12 \pm \sqrt{144 - 144}}{18} = \frac{12 \pm 0}{18} = \frac{12}{18} = \frac{2}{3}$

(6) $x^2 - 2\sqrt{3}x + 3 = 0$

$a = 1$, $b = -2\sqrt{3}$, $c = 3$

$x = \frac{2\sqrt{3} \pm \sqrt{(-2\sqrt{3})^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{2\sqrt{3} \pm \sqrt{12 - 12}}{2} = \frac{2\sqrt{3} \pm 0}{2} = \sqrt{3}$

3. 最終的な答え

(1) $x = \frac{-7 \pm \sqrt{37}}{6}$

(2) $x = \frac{3 \pm \sqrt{17}}{2}$

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

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

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

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