与えられた6つの2次式を平方完成させる問題です。 (1) $3x^2 - 12x$ (2) $2x^2 + 4x + 1$ (3) $3x^2 - 9x + 7$ (4) $-2x^2 - 8x - 10$ (5) $-x^2 + 3x - 2$ (6) $-2x^2 - 10x + 3$

1. 問題の内容

与えられた6つの2次式を平方完成させる問題です。

(1) $3x^2 - 12x$

(2) $2x^2 + 4x + 1$

(3) $3x^2 - 9x + 7$

(4) $-2x^2 - 8x - 10$

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

(6) $-2x^2 - 10x + 3$

2. 解き方の手順

(1) $3x^2 - 12x$

$3(x^2 - 4x) = 3(x^2 - 4x + 4 - 4) = 3((x - 2)^2 - 4) = 3(x - 2)^2 - 12$

(2) $2x^2 + 4x + 1$

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

(3) $3x^2 - 9x + 7$

$3(x^2 - 3x) + 7 = 3(x^2 - 3x + (\frac{3}{2})^2 - (\frac{3}{2})^2) + 7 = 3((x - \frac{3}{2})^2 - \frac{9}{4}) + 7 = 3(x - \frac{3}{2})^2 - \frac{27}{4} + \frac{28}{4} = 3(x - \frac{3}{2})^2 + \frac{1}{4}$

(4) $-2x^2 - 8x - 10$

$-2(x^2 + 4x) - 10 = -2(x^2 + 4x + 4 - 4) - 10 = -2((x + 2)^2 - 4) - 10 = -2(x + 2)^2 + 8 - 10 = -2(x + 2)^2 - 2$

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

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

(6) $-2x^2 - 10x + 3$

$-2(x^2 + 5x) + 3 = -2(x^2 + 5x + (\frac{5}{2})^2 - (\frac{5}{2})^2) + 3 = -2((x + \frac{5}{2})^2 - \frac{25}{4}) + 3 = -2(x + \frac{5}{2})^2 + \frac{25}{2} + \frac{6}{2} = -2(x + \frac{5}{2})^2 + \frac{31}{2}$

3. 最終的な答え

(1) $3(x - 2)^2 - 12$

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

(3) $3(x - \frac{3}{2})^2 + \frac{1}{4}$

(4) $-2(x + 2)^2 - 2$

(5) $-(x - \frac{3}{2})^2 + \frac{1}{4}$

(6) $-2(x + \frac{5}{2})^2 + \frac{31}{2}$