Question

(4x)2+4x+1=0能否用因式分解解

Answer 1

4x" + 4x + 1 = 0
( 2x )" + 2( 2x ) + 1 = 0
( 2x + 1 )" = 0
解方程
x1 = x2 = - 1/2

如果真是 ( 4x )" + 4x + 1 = 0 ，
实际就是 16x" + 4x + 1 = 0 ，
b" - 4ac = 16 - 64 < 0 ，
方程就没有实数解了，可是在复数范围，
仍然是可以配方分解因式解方程的

4( 4x )" + 16x + 4 = 0
( 8x )" + 2( 8x ) + 1" - 1 + 4 = 0
( 8x + 1 )" + 3 = 0
( 8x + 1 )" - ( -3 ) = 0
( 8x + 1 )" - ( √3i )" = 0
( 8x + 1 + √3i )( 8x + 1 - √3i ) = 0

方程的复数解，就是
x1 = ( -1 -√3i ) / 8 = -1/8 - ( √3 / 8 )i ，
x2 = ( -1 + √3i ) / 8 = -1/8 + ( √3 / 8 )i

Answer 2

不能，不能