与えられた6つの二次式を平方完成させる問題です。

1. 問題の内容

与えられた6つの二次式を平方完成させる問題です。

2. 解き方の手順

(1) $x^2 + 8x$

$x^2 + 8x = (x + 4)^2 - 4^2 = (x + 4)^2 - 16$

(2) $x^2 - 6x + 8$

$x^2 - 6x + 8 = (x - 3)^2 - 3^2 + 8 = (x - 3)^2 - 9 + 8 = (x - 3)^2 - 1$

(3) $2x^2 - 8x + 5$

$2x^2 - 8x + 5 = 2(x^2 - 4x) + 5 = 2((x - 2)^2 - 2^2) + 5 = 2(x - 2)^2 - 8 + 5 = 2(x - 2)^2 - 3$

(4) $3x^2 + 6x + 2$

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

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

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

(6) $-2x^2 + 6x + 4$

$-2x^2 + 6x + 4 = -2(x^2 - 3x) + 4 = -2((x - \frac{3}{2})^2 - (\frac{3}{2})^2) + 4 = -2(x - \frac{3}{2})^2 + 2 \cdot \frac{9}{4} + 4 = -2(x - \frac{3}{2})^2 + \frac{9}{2} + \frac{8}{2} = -2(x - \frac{3}{2})^2 + \frac{17}{2}$

3. 最終的な答え

(1) $(x + 4)^2 - 16$

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

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

(4) $3(x + 1)^2 - 1$

(5) $(x + \frac{1}{2})^2 - \frac{9}{4}$

(6) $-2(x - \frac{3}{2})^2 + \frac{17}{2}$