## 1. 問題の内容

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1. 問題の内容

次の2次方程式を解け。

(1) $x^2 = -9$

(2) $x^2 + 3x + 10 = 0$

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

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

(5) $(\sqrt{2}-1)x^2 + \sqrt{2}x + 1 = 0$

(6) $(x+1)(x+3) = x(9-2x)$

(7) $1.4x - 1.2x^2 = 0.6$

(8) $\frac{x^2+1}{2} = \frac{x-1}{3}$

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2. 解き方の手順

(1) $x^2 = -9$

$x = \pm \sqrt{-9}$

$x = \pm \sqrt{9}i$

$x = \pm 3i$

(2) $x^2 + 3x + 10 = 0$

解の公式を用いる。$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(10)}}{2(1)}$

$x = \frac{-3 \pm \sqrt{9 - 40}}{2}$

$x = \frac{-3 \pm \sqrt{-31}}{2}$

$x = \frac{-3 \pm \sqrt{31}i}{2}$

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

解の公式を用いる。$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$x = \frac{4 \pm \sqrt{16 - 32}}{2}$

$x = \frac{4 \pm \sqrt{-16}}{2}$

$x = \frac{4 \pm 4i}{2}$

$x = 2 \pm 2i$

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

$t = x-1$ とおく。

$2t^2 + 2t + 1 = 0$

解の公式を用いる。$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$t = \frac{-2 \pm \sqrt{4 - 8}}{4}$

$t = \frac{-2 \pm \sqrt{-4}}{4}$

$t = \frac{-2 \pm 2i}{4}$

$t = \frac{-1 \pm i}{2}$

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

$x = 1 + \frac{-1 \pm i}{2}$

$x = \frac{2}{2} + \frac{-1 \pm i}{2}$

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

(5) $(\sqrt{2}-1)x^2 + \sqrt{2}x + 1 = 0$

解の公式を用いる。$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

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

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

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

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

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

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

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

$x = \frac{-\sqrt{2} + 2 - \sqrt{2}}{2(\sqrt{2}-1)} = \frac{2-2\sqrt{2}}{2(\sqrt{2}-1)} = \frac{2(1-\sqrt{2})}{2(\sqrt{2}-1)} = -1$

$x = \frac{-\sqrt{2} - 2 + \sqrt{2}}{2(\sqrt{2}-1)} = \frac{-2}{2(\sqrt{2}-1)} = \frac{-1}{\sqrt{2}-1} = \frac{-1(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{-\sqrt{2}-1}{2-1} = -\sqrt{2}-1$

(6) $(x+1)(x+3) = x(9-2x)$

$x^2 + 4x + 3 = 9x - 2x^2$

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

解の公式を用いる。$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$x = \frac{5 \pm \sqrt{25 - 36}}{6}$

$x = \frac{5 \pm \sqrt{-11}}{6}$

$x = \frac{5 \pm \sqrt{11}i}{6}$

(7) $1.4x - 1.2x^2 = 0.6$

$1.2x^2 - 1.4x + 0.6 = 0$

$12x^2 - 14x + 6 = 0$

$6x^2 - 7x + 3 = 0$

解の公式を用いる。$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{7 \pm \sqrt{(-7)^2 - 4(6)(3)}}{2(6)}$

$x = \frac{7 \pm \sqrt{49 - 72}}{12}$

$x = \frac{7 \pm \sqrt{-23}}{12}$

$x = \frac{7 \pm \sqrt{23}i}{12}$

(8) $\frac{x^2+1}{2} = \frac{x-1}{3}$

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

$3x^2 + 3 = 2x - 2$

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

解の公式を用いる。$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$x = \frac{2 \pm \sqrt{4 - 60}}{6}$

$x = \frac{2 \pm \sqrt{-56}}{6}$

$x = \frac{2 \pm \sqrt{56}i}{6}$

$x = \frac{2 \pm 2\sqrt{14}i}{6}$

$x = \frac{1 \pm \sqrt{14}i}{3}$

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3. 最終的な答え

(1) $x = \pm 3i$

(2) $x = \frac{-3 \pm \sqrt{31}i}{2}$

(3) $x = 2 \pm 2i$

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

(5) $x = -1, -\sqrt{2}-1$

(6) $x = \frac{5 \pm \sqrt{11}i}{6}$

(7) $x = \frac{7 \pm \sqrt{23}i}{12}$

(8) $x = \frac{1 \pm \sqrt{14}i}{3}$