【对应训练】
4.方程$x^{2}+2x= 3$的根是 ()
A.$x_{1}= 1,x_{2}= -3$
B.$x_{1}= -1,x_{2}= 3$
C.$x_{1}= -1+\sqrt {3},x_{2}= -1-\sqrt {3}$
D.$x_{1}= 1+\sqrt {3},x_{2}= 1-\sqrt {3}$

A

5.用配方法解方程$x^{2}-4x-5= 0$,则$x^{2}-4x+$____$=5+$____,所以$x_{1}= $____,$x_{2}= $____.

4 4 5 -1

6.解方程:
(1)$x^{2}-6x= -5$;
(2)$2x^{2}-3x-3= 0$;
(3)$y(y-4)= -1-2y$;
(4)$x^{2}+px+q= 0(p^{2}＞4q)$.

(1) $x_{1}=5,x_{2}=1$ (2) $x_{1}=\frac{3+\sqrt{33}}{4},x_{2}=\frac{3-\sqrt{33}}{4}$ (3) $y_{1}=y_{2}=1$ (4) $x_{1}=\frac{-p+\sqrt{p^{2}-4q}}{2},x_{2}=\frac{-p-\sqrt{p^{2}-4q}}{2}$

7.把方程$x^{2}-3x+p= 0$配方后,得到$(x+m)^{2}= \frac {1}{2}$.
(1)求常数$p与m$的值;
(2)求此方程的根.

(1) $\because x^{2}-3x+p=0,\therefore x^{2}-3x=-p,x^{2}-3x+(\frac{3}{2})^{2}=-p+(\frac{3}{2})^{2},(x-\frac{3}{2})^{2}=-p+\frac{9}{4},\therefore m=-\frac{3}{2},-p+\frac{9}{4}=\frac{1}{2}$，解得 $p=\frac{7}{4},m=-\frac{3}{2}$ (2) $\because x^{2}-3x+p=0,\therefore (x-\frac{3}{2})^{2}=\frac{1}{2},x-\frac{3}{2}=\pm\frac{\sqrt{2}}{2}$，即方程的根是 $x_{1}=\frac{3+\sqrt{2}}{2},x_{2}=\frac{3-\sqrt{2}}{2}$