-3x的平方-4x+4=0用公式法解

2024-11-15 16:34:55
推荐回答(3个)
回答(1):

-3x^2-4x+4=0
x1=(4+(4^2-4*(-3)*4)^0.5)/(2*(-3))
=(4+(16+48)^0.5)/(-6)
=(4+64^0.5)/(-6)
=(4+8)/(-6)
=-2
x2=(4-(4^2-4*(-3)*4)^0.5)/(2*(-3))
=(4-(16+48)^0.5)/(-6)
=(4-64^0.5)/(-6)
=(4-8)/(-6)
=2/3

回答(2):

a=-3 b=-4 c=4
b^2-4ac=16-4*(-3)*4=64
x1=(4+8)/-6=-2
x2=(4-8)/-6=2/3

回答(3):

-3[x^2 + (4/3)x + (2/3)^2 - (2/3)^2]=-4
[x + (2/3)]^2 - (2/3)^2=4/3
[x + (2/3)]^2=16/9
x + (2/3)=±4/3
x=2/3或x=-2