与えられた2次方程式を解く問題です。

はい、承知いたしました。画像に写っているすべての方程式を解きます。

1. 問題の内容

与えられた2次方程式を解く問題です。

2. 解き方の手順

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

$x^2 = 5$

$x = \pm \sqrt{5}$

(2) $x^2 - 27 = 0$

$x^2 = 27$

$x = \pm \sqrt{27} = \pm 3\sqrt{3}$

(3) $x^2 - 81 = 0$

$x^2 = 81$

$x = \pm 9$

(4) $9x^2 - 45 = 0$

$9x^2 = 45$

$x^2 = 5$

$x = \pm \sqrt{5}$

(5) $4x^2 - 48 = 0$

$4x^2 = 48$

$x^2 = 12$

$x = \pm \sqrt{12} = \pm 2\sqrt{3}$

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

$3x^2 = 8$

$x^2 = \frac{8}{3}$

$x = \pm \sqrt{\frac{8}{3}} = \pm \frac{2\sqrt{2}}{\sqrt{3}} = \pm \frac{2\sqrt{6}}{3}$

(7) $6x^2 - 128 = 0$

$6x^2 = 128$

$x^2 = \frac{128}{6} = \frac{64}{3}$

$x = \pm \sqrt{\frac{64}{3}} = \pm \frac{8}{\sqrt{3}} = \pm \frac{8\sqrt{3}}{3}$

(8) $4x^2 - 7 = 0$

$4x^2 = 7$

$x^2 = \frac{7}{4}$

$x = \pm \sqrt{\frac{7}{4}} = \pm \frac{\sqrt{7}}{2}$

(9) $4x^2 - 9 = 0$

$4x^2 = 9$

$x^2 = \frac{9}{4}$

$x = \pm \sqrt{\frac{9}{4}} = \pm \frac{3}{2}$

(10) $(x+8)^2 - 27 = 0$

$(x+8)^2 = 27$

$x+8 = \pm \sqrt{27} = \pm 3\sqrt{3}$

$x = -8 \pm 3\sqrt{3}$

(11) $(x-7)^2 - 12 = 0$

$(x-7)^2 = 12$

$x-7 = \pm \sqrt{12} = \pm 2\sqrt{3}$

$x = 7 \pm 2\sqrt{3}$

(12) $(x+2)^2 - 50 = 0$

$(x+2)^2 = 50$

$x+2 = \pm \sqrt{50} = \pm 5\sqrt{2}$

$x = -2 \pm 5\sqrt{2}$

(13) $(x+3)^2 = 25$

$x+3 = \pm \sqrt{25} = \pm 5$

$x = -3 \pm 5$

$x = -3 + 5 = 2$

$x = -3 - 5 = -8$

$x = 2, -8$

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

$(x+5)^2 = 1$

$x+5 = \pm \sqrt{1} = \pm 1$

$x = -5 \pm 1$

$x = -5 + 1 = -4$

$x = -5 - 1 = -6$

$x = -4, -6$

(15) $(x-5)^2 - 64 = 0$

$(x-5)^2 = 64$

$x-5 = \pm \sqrt{64} = \pm 8$

$x = 5 \pm 8$

$x = 5 + 8 = 13$

$x = 5 - 8 = -3$

$x = 13, -3$

(16) $2(x-4)^2 - 12 = 0$

$2(x-4)^2 = 12$

$(x-4)^2 = 6$

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

$x = 4 \pm \sqrt{6}$

(17) $3(x+5)^2 - 24 = 0$

$3(x+5)^2 = 24$

$(x+5)^2 = 8$

$x+5 = \pm \sqrt{8} = \pm 2\sqrt{2}$

$x = -5 \pm 2\sqrt{2}$

(18) $2(x+3)^2 - 54 = 0$

$2(x+3)^2 = 54$

$(x+3)^2 = 27$

$x+3 = \pm \sqrt{27} = \pm 3\sqrt{3}$

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

(19) $2(x+4)^2 - 18 = 0$

$2(x+4)^2 = 18$

$(x+4)^2 = 9$

$x+4 = \pm \sqrt{9} = \pm 3$

$x = -4 \pm 3$

$x = -4 + 3 = -1$

$x = -4 - 3 = -7$

$x = -1, -7$

(20) $3(x+1)^2 - 3 = 0$

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

$(x+1)^2 = 1$

$x+1 = \pm \sqrt{1} = \pm 1$

$x = -1 \pm 1$

$x = -1 + 1 = 0$

$x = -1 - 1 = -2$

$x = 0, -2$

(21) $4(x-1)^2 - 144 = 0$

$4(x-1)^2 = 144$

$(x-1)^2 = 36$

$x-1 = \pm \sqrt{36} = \pm 6$

$x = 1 \pm 6$

$x = 1 + 6 = 7$

$x = 1 - 6 = -5$

$x = 7, -5$

3. 最終的な答え

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

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

(3) $x = \pm 9$

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

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

(6) $x = \pm \frac{2\sqrt{6}}{3}$

(7) $x = \pm \frac{8\sqrt{3}}{3}$

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

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

(10) $x = -8 \pm 3\sqrt{3}$

(11) $x = 7 \pm 2\sqrt{3}$

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

(13) $x = 2, -8$

(14) $x = -4, -6$

(15) $x = 13, -3$

(16) $x = 4 \pm \sqrt{6}$

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

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

(19) $x = -1, -7$

(20) $x = 0, -2$

(21) $x = 7, -5$