与えられた9つの2次方程式を解く問題です。

1. 問題の内容

与えられた9つの2次方程式を解く問題です。

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られます。解の公式は次の通りです。

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

各方程式について、$a, b, c$ の値を特定し、解の公式に代入して解を求めます。

(1) $x^2 + x + 5 = 0$

$a = 1, b = 1, c = 5$

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(5)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 20}}{2} = \frac{-1 \pm \sqrt{-19}}{2} = \frac{-1 \pm i\sqrt{19}}{2}$

(2) $x^2 - 3x + 9 = 0$

$a = 1, b = -3, c = 9$

$x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(9)}}{2(1)} = \frac{3 \pm \sqrt{9 - 36}}{2} = \frac{3 \pm \sqrt{-27}}{2} = \frac{3 \pm 3i\sqrt{3}}{2}$

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

$a = 3, b = 2, c = -5$

$x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-5)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 60}}{6} = \frac{-2 \pm \sqrt{64}}{6} = \frac{-2 \pm 8}{6}$

$x = \frac{-2 + 8}{6} = \frac{6}{6} = 1$

$x = \frac{-2 - 8}{6} = \frac{-10}{6} = -\frac{5}{3}$

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

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

$x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}$

(5) $2x^2 + 4x + 3 = 0$

$a = 2, b = 4, c = 3$

$x = \frac{-4 \pm \sqrt{4^2 - 4(2)(3)}}{2(2)} = \frac{-4 \pm \sqrt{16 - 24}}{4} = \frac{-4 \pm \sqrt{-8}}{4} = \frac{-4 \pm 2i\sqrt{2}}{4} = -1 \pm \frac{i\sqrt{2}}{2}$

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

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

$x = \frac{-2 \pm \sqrt{2^2 - 4(-3)(2)}}{2(-3)} = \frac{-2 \pm \sqrt{4 + 24}}{-6} = \frac{-2 \pm \sqrt{28}}{-6} = \frac{-2 \pm 2\sqrt{7}}{-6} = \frac{1 \mp \sqrt{7}}{3}$

(7) $5x^2 + 3x + 2 = 0$

$a = 5, b = 3, c = 2$

$x = \frac{-3 \pm \sqrt{3^2 - 4(5)(2)}}{2(5)} = \frac{-3 \pm \sqrt{9 - 40}}{10} = \frac{-3 \pm \sqrt{-31}}{10} = \frac{-3 \pm i\sqrt{31}}{10}$

(8) $x^2 - \sqrt{5}x + 2 = 0$

$a = 1, b = -\sqrt{5}, c = 2$

$x = \frac{\sqrt{5} \pm \sqrt{(-\sqrt{5})^2 - 4(1)(2)}}{2(1)} = \frac{\sqrt{5} \pm \sqrt{5 - 8}}{2} = \frac{\sqrt{5} \pm \sqrt{-3}}{2} = \frac{\sqrt{5} \pm i\sqrt{3}}{2}$

(9) $2x^2 + 4\sqrt{3}x + 7 = 0$

$a = 2, b = 4\sqrt{3}, c = 7$

$x = \frac{-4\sqrt{3} \pm \sqrt{(4\sqrt{3})^2 - 4(2)(7)}}{2(2)} = \frac{-4\sqrt{3} \pm \sqrt{48 - 56}}{4} = \frac{-4\sqrt{3} \pm \sqrt{-8}}{4} = \frac{-4\sqrt{3} \pm 2i\sqrt{2}}{4} = -\sqrt{3} \pm \frac{i\sqrt{2}}{2}$

3. 最終的な答え

(1) $x = \frac{-1 \pm i\sqrt{19}}{2}$

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

(3) $x = 1, -\frac{5}{3}$

(4) $x = \frac{3 \pm \sqrt{13}}{2}$

(5) $x = -1 \pm \frac{i\sqrt{2}}{2}$

(6) $x = \frac{1 \mp \sqrt{7}}{3}$

(7) $x = \frac{-3 \pm i\sqrt{31}}{10}$

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

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