一題簡單math 題

sol
(x-2)^2+(0-5)^2=13
移项得 (x-2)^2=-12
x-2=± sqrt12 i
=±2sqrt3 i

是否打錯

(X-2)^2 +(0-5)^2 = 13
定
(X-2)^2 +(X - 5)^2 = 13

(X-2)^2 +(0-5)^2 = 13
x^2-4x+4+(-5)^2=13
x^2-4x+4+25-13=0
x^2-4x+16=0

2011-04-16 15:42:37 補充：
如果要搵X,你可以继續:

x^2-4x+16=0
(x^2-4x+4^2-4^2)+16=0
(x-4)^2-16+16=0
(x-1)^2 =0
x-1=0
x=1

*此乃completing the square method

(X-2)^2 +(0-5)^2 = 13
x^2-4x+4+25=13
x^2-4x+29=13
x^2-4x+16=0
(x^2-40+16=0
(x^2-4x+(-4/2)^2-(-4/2)^2)+16=0
(x-2)^2+16-(-4)^2/4(1)=0
(x-2)^2+16-4=0
(x-2)^2+12=0

2011-04-15 22:42:18 補充：
有可能.......

2011-04-15 22:44:01 補充：
打多左行>.<

(X-2)^2 +(0-5)^2 = 13
x^2-4x+4+25=13
x^2-4x+29=13
x^2-4x+16=0
(x^2-4x+(-4/2)^2-(-4/2)^2)+16=0
(x-2)^2+16-(-4)^2/4(1)=0
(x-2)^2+16-4=0
(x-2)^2+12=0