【例1】解下列方程:
(1)$x^2 - 25 = 0$;
(2)$4x^2 = 0$;
(3)$0.8x^2 - 4 = 0$;
(4)$4.3 - 6x^2 = 2.8$。
(1)$x^2 - 25 = 0$;
(2)$4x^2 = 0$;
(3)$0.8x^2 - 4 = 0$;
(4)$4.3 - 6x^2 = 2.8$。
答案
(1)移项,得 $x^2=25$,
$\therefore x=\pm5$,即 $x_1=5,x_2=-5$.
(2)将二次项系数化为 1,得 $x^2=0$,
$\therefore x=0$,即 $x_1=x_2=0$.
(3)移项,并将二次项系数化为 1,得 $x^2=5$,
$\therefore x=\pm\sqrt{5}$,即 $x_1=\sqrt{5},x_2=-\sqrt{5}$.
(4)移项,并将二次项系数化为 1,得 $x^2=\frac{1}{4}$,
$\therefore x=\pm\frac{1}{2}$,即 $x_1=\frac{1}{2},x_2=-\frac{1}{2}$.
$\therefore x=\pm5$,即 $x_1=5,x_2=-5$.
(2)将二次项系数化为 1,得 $x^2=0$,
$\therefore x=0$,即 $x_1=x_2=0$.
(3)移项,并将二次项系数化为 1,得 $x^2=5$,
$\therefore x=\pm\sqrt{5}$,即 $x_1=\sqrt{5},x_2=-\sqrt{5}$.
(4)移项,并将二次项系数化为 1,得 $x^2=\frac{1}{4}$,
$\therefore x=\pm\frac{1}{2}$,即 $x_1=\frac{1}{2},x_2=-\frac{1}{2}$.
【变式】解下列方程:
(1)$5x^2=125$;
(2)$(x+3)(x-3)=9$。
(1)$5x^2=125$;
(2)$(x+3)(x-3)=9$。
答案
(1)移项,并将二次项系数化为 1,得 $x^2=25$,
$\therefore x=\pm5$,即 $x_1=5,x_2=-5$.
(2)方程化为一般形式为 $x^2=18$,
$\therefore x=\pm3\sqrt{2}$,即 $x_1=3\sqrt{2},x_2=-3\sqrt{2}$.
$\therefore x=\pm5$,即 $x_1=5,x_2=-5$.
(2)方程化为一般形式为 $x^2=18$,
$\therefore x=\pm3\sqrt{2}$,即 $x_1=3\sqrt{2},x_2=-3\sqrt{2}$.
【例2】解下列方程:
(1)$(\sqrt{2}x - 2)^2 = 6$;
(2)$3(x - 1)^2 - 6 = 0$;
(3)$(x + \sqrt{2})^2 = (1 + \sqrt{2})^2$。
(1)$(\sqrt{2}x - 2)^2 = 6$;
(2)$3(x - 1)^2 - 6 = 0$;
(3)$(x + \sqrt{2})^2 = (1 + \sqrt{2})^2$。
答案
(1)根据平方根的意义,得$\sqrt{2}x-2=\pm\sqrt{6}$,
$\therefore x_1=\sqrt{2}+\sqrt{3},x_2=\sqrt{2}-\sqrt{3}$.
(2)移项,并将二次项系数化为 1,得$(x-1)^2=2$,
$\therefore x-1=\pm\sqrt{2}$,$\therefore x_1=1+\sqrt{2},x_2=1-\sqrt{2}$.
(3)根据平方根的意义,得 $x+\sqrt{2}=1+\sqrt{2}$,或 $x+\sqrt{2}=-1-\sqrt{2}$,
$\therefore x_1=1,x_2=-1-2\sqrt{2}$.
$\therefore x_1=\sqrt{2}+\sqrt{3},x_2=\sqrt{2}-\sqrt{3}$.
(2)移项,并将二次项系数化为 1,得$(x-1)^2=2$,
$\therefore x-1=\pm\sqrt{2}$,$\therefore x_1=1+\sqrt{2},x_2=1-\sqrt{2}$.
(3)根据平方根的意义,得 $x+\sqrt{2}=1+\sqrt{2}$,或 $x+\sqrt{2}=-1-\sqrt{2}$,
$\therefore x_1=1,x_2=-1-2\sqrt{2}$.
【变式】解下列方程:
(1)$6(x-1)^2 -54=0$;
(2)$x^2 +4x +4=1$。
(1)$6(x-1)^2 -54=0$;
(2)$x^2 +4x +4=1$。
答案
(1)移项,并将二次项系数化为 1,得$(x-1)^2=9$,
$\therefore x-1=\pm3$,$\therefore x_1=4,x_2=-2$.
(2)由 $x^2+4x+4=1$,得$(x+2)^2=1$,
$\therefore x+2=\pm1$,$\therefore x_1=-1,x_2=-3$.
$\therefore x-1=\pm3$,$\therefore x_1=4,x_2=-2$.
(2)由 $x^2+4x+4=1$,得$(x+2)^2=1$,
$\therefore x+2=\pm1$,$\therefore x_1=-1,x_2=-3$.
登录