与えられた6つの二次方程式を、解の公式を使って解きます。 (1) $x^2 - x - 1 = 0$ (2) $x^2 + 4x - 8 = 0$ (3) $x^2 - 6x - 3 = 0$ (4) $2x^2 - 5x + 3 = 0$ (5) $2x^2 = x + 4$ (6) $3x^2 = 2x + 1$

1. 問題の内容

与えられた6つの二次方程式を、解の公式を使って解きます。

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

(2) $x^2 + 4x - 8 = 0$

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

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

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

(6) $3x^2 = 2x + 1$

2. 解き方の手順

二次方程式 $ax^2 + bx + c = 0$ の解の公式は次のとおりです。

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

それぞれの問題について、$a$, $b$, $c$ を特定し、解の公式に代入して解を求めます。

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

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

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

(2) $x^2 + 4x - 8 = 0$

$a = 1$, $b = 4$, $c = -8$

$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-8)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 32}}{2} = \frac{-4 \pm \sqrt{48}}{2} = \frac{-4 \pm 4\sqrt{3}}{2} = -2 \pm 2\sqrt{3}$

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

$a = 1$, $b = -6$, $c = -3$

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-3)}}{2(1)} = \frac{6 \pm \sqrt{36 + 12}}{2} = \frac{6 \pm \sqrt{48}}{2} = \frac{6 \pm 4\sqrt{3}}{2} = 3 \pm 2\sqrt{3}$

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

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

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(3)}}{2(2)} = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm \sqrt{1}}{4} = \frac{5 \pm 1}{4}$

$x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}$

$x = \frac{5 - 1}{4} = \frac{4}{4} = 1$

(5) $2x^2 = x + 4 \Rightarrow 2x^2 - x - 4 = 0$

$a = 2$, $b = -1$, $c = -4$

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-4)}}{2(2)} = \frac{1 \pm \sqrt{1 + 32}}{4} = \frac{1 \pm \sqrt{33}}{4}$

(6) $3x^2 = 2x + 1 \Rightarrow 3x^2 - 2x - 1 = 0$

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

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6}$

$x = \frac{2 + 4}{6} = \frac{6}{6} = 1$

$x = \frac{2 - 4}{6} = \frac{-2}{6} = -\frac{1}{3}$

3. 最終的な答え

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

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

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

(4) $x = 1, \frac{3}{2}$

(5) $x = \frac{1 \pm \sqrt{33}}{4}$

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