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4

2

$\frac {9}{4}$

$\frac {3}{2}$

$\frac {9}{16}$

$\frac {3}{4}$

6

3

B

 解：$x^{2}+2x+1=4$

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

 $x_1=1，$$x_2=-3$

 解：$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$

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

 $x_{1}=1，$$x_{2}=2$

 解：$(x+3)^{2}=0$

 $\ \ \ \ \ \ \ x=-3$

 解：$x^{2}+\frac{3}{4}x+\frac{9}{64}=1+\frac{9}{64}$

 $\ \ \ \ \ \ \ (x+\frac{3}{8})^{2}=\frac{73}{64}$

 $x_1=\frac{\sqrt{73}}{8}-\frac{3}{8}，$$x_2=-\frac{\sqrt{73}}{8}-\frac{3}{8}$

 解：$x^2 - x - 6 = 0$$(x - 3)(x + 2) = 0$$x - 3 = 0$或$x + 2 = 0$$x_1 = 3，$$x_2 = -2$

 解：$x^2 + 4x + 5 = 0$$\Delta = 4^2 - 4×1×5 = 16 - 20 = -4 ＜ 0$方程無(wú)實(shí)數(shù)根

 解：$x^2 - 2x - 2 = 0$$x^2 - 2x = 2$$x^2 - 2x + 1 = 2 + 1$$(x - 1)^2 = 3$$x - 1 = ±\sqrt{3}$$x_1 = 1 + \sqrt{3}，$$x_2 = 1 - \sqrt{3}$

 解：$p^2 + 5p - 1 = 0$$p^2 + 5p = 1$$p^2 + 5p + (\frac{5}{2})^2 = 1 + (\frac{5}{2})^2$$(p + \frac{5}{2})^2 = \frac{29}{4}$$p + \frac{5}{2} = ±\frac{\sqrt{29}}{2}$$p_1 = \frac{-5 + \sqrt{29}}{2}，$$p_2 = \frac{-5 - \sqrt{29}}{2}$

(1)解：$x^2 - x - 6 = 0$
$(x - 3)(x + 2) = 0$
$x - 3 = 0$或$x + 2 = 0$
$x_1 = 3$，$x_2 = -2$
(2)解：$x^2 + 4x + 5 = 0$
$\Delta = 4^2 - 4×1×5 = 16 - 20 = -4 ＜ 0$
方程無(wú)實(shí)數(shù)根
(3)解：$x^2 - 2x - 2 = 0$
$x^2 - 2x = 2$
$x^2 - 2x + 1 = 2 + 1$
$(x - 1)^2 = 3$
$x - 1 = ±\sqrt{3}$
$x_1 = 1 + \sqrt{3}$，$x_2 = 1 - \sqrt{3}$
(4)解：$p^2 + 5p - 1 = 0$
$p^2 + 5p = 1$
$p^2 + 5p + (\frac{5}{2})^2 = 1 + (\frac{5}{2})^2$
$(p + \frac{5}{2})^2 = \frac{29}{4}$
$p + \frac{5}{2} = ±\frac{\sqrt{29}}{2}$
$p_1 = \frac{-5 + \sqrt{29}}{2}$，$p_2 = \frac{-5 - \sqrt{29}}{2}$

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