1. 方程$2x^2=6$的根是 【 】
A.$x_1=x_2=\sqrt{3}$
B.$x_1=\sqrt{3}, x_2=-\sqrt{3}$
C.$x=\sqrt{3}$
D.$x=-\sqrt{3}$
A.$x_1=x_2=\sqrt{3}$
B.$x_1=\sqrt{3}, x_2=-\sqrt{3}$
C.$x=\sqrt{3}$
D.$x=-\sqrt{3}$
答案
B
2. 下列方程不能用直接开平方法求解的是 【 】
A.$x^2 - 4 = 0$
B.$(x - 1)^2 - 9 = 0$
C.$x^2 + 3x = 0$
D.$(x - 1)^2 = (2x + 1)^2$
A.$x^2 - 4 = 0$
B.$(x - 1)^2 - 9 = 0$
C.$x^2 + 3x = 0$
D.$(x - 1)^2 = (2x + 1)^2$
答案
C
3. 如图是一个数值运算程序图,若输出的结果为-12,则输入x的值为 【 】

A.-1
B.3
C.-1或3
D.无法确定
A.-1
B.3
C.-1或3
D.无法确定
答案
C
4. 一元二次方程$x^2 - 8 = 0$的解为
$x_1=2\sqrt{2}, x_2=-2\sqrt{2}$
。答案
4. $x_1=2\sqrt{2}, x_2=-2\sqrt{2}$
5. 一元二次方程$(x - 2)^2 = 0$的根是
$x_1=x_2=2$
。答案
5. $x_1=x_2=2$
6. 若关于$x$的方程$(mx + 1)^2 - 9 = 0$的一个根为1,则$m$的值为
-4或2
.答案
6. -4或2
7.若$(a^2 + b^2 - 3)^2 = 25$,则$a^2 + b^2 =$
8
.答案
7. 8
8. 用直接开平方法解方程:
(1) $(x + 1)^2 = 4$;
(2) $9(x + 2)^2 - 16 = 0$;
(3) $(x + 3)^2 = 2$;
(4) $(x - 1)^2 - 12 = 0$;
(5) $4(2x + 1)^2 - 25 = 0$;
(6) $(x - 3)^2 = 4(3x - 1)^2$。
(1) $(x + 1)^2 = 4$;
(2) $9(x + 2)^2 - 16 = 0$;
(3) $(x + 3)^2 = 2$;
(4) $(x - 1)^2 - 12 = 0$;
(5) $4(2x + 1)^2 - 25 = 0$;
(6) $(x - 3)^2 = 4(3x - 1)^2$。
答案
8. (1) 直接开平方,得$x + 1 = \pm 2. \therefore x_1 = 1, x_2 = -3$
(2) 原方程可化为$(x + 2)^2 = \frac{16}{9}$. 直接开平方,得$x + 2 = \pm \frac{4}{3}$. $\therefore x_1 = -\frac{2}{3}, x_2 = -\frac{10}{3}$
(3) 直接开平方,得$x + 3 = \pm \sqrt{2}. \therefore x_1 = -3 + \sqrt{2}, x_2 = -3 - \sqrt{2}$
(4) 移项,得$(x - 1)^2 = 12$. 直接开平方,得$x - 1 = \pm 2\sqrt{3}. \therefore x_1 = 1 + 2\sqrt{3}, x_2 = 1 - 2\sqrt{3}$
(5) 原方程可化为$(2x + 1)^2 = \frac{25}{4}$. 直接开平方,得$2x + 1 = \pm \frac{5}{2}. \therefore x_1 = \frac{3}{4}, x_2 = -\frac{7}{4}$
(6) 直接开平方,得$x - 3 = \pm 2(3x - 1). \therefore x_1 = -\frac{1}{5}, x_2 = \frac{5}{7}$
(2) 原方程可化为$(x + 2)^2 = \frac{16}{9}$. 直接开平方,得$x + 2 = \pm \frac{4}{3}$. $\therefore x_1 = -\frac{2}{3}, x_2 = -\frac{10}{3}$
(3) 直接开平方,得$x + 3 = \pm \sqrt{2}. \therefore x_1 = -3 + \sqrt{2}, x_2 = -3 - \sqrt{2}$
(4) 移项,得$(x - 1)^2 = 12$. 直接开平方,得$x - 1 = \pm 2\sqrt{3}. \therefore x_1 = 1 + 2\sqrt{3}, x_2 = 1 - 2\sqrt{3}$
(5) 原方程可化为$(2x + 1)^2 = \frac{25}{4}$. 直接开平方,得$2x + 1 = \pm \frac{5}{2}. \therefore x_1 = \frac{3}{4}, x_2 = -\frac{7}{4}$
(6) 直接开平方,得$x - 3 = \pm 2(3x - 1). \therefore x_1 = -\frac{1}{5}, x_2 = \frac{5}{7}$
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