solve by completing square
Re: solve by completing square
2x²-3x=1
so 2x²-3x-1=0 ------ let A
let ax²+bx+c=0
hence x=(-b±?(b²-4ac))/2a ------ let B
here from A & B
a=2, b=-3, c=-1
so x=[-(-3)±?((-3)²-4.2.-1)]/2.2
hence x=[3±?(9+8)]/4
x=[3±?17]/4
hence x1=[3+?17]/4 or x2=[3-?17]/4
x1=[3+4.123]/4 or x2=[3-4.123]/4
Re: solve by completing square
Quote:
2x²-3x=1
so 2x²-3x-1=0 ------ let A
let ax²+bx+c=0
hence x=(-b±?(b²-4ac))/2a ------ let B
here from A & B
a=2, b=-3, c=-1
so x=[-(-3)±?((-3)²-4.2.-1)]/2.2
hence x=[3±?(9+8)]/4
x=[3±?17]/4
hence x1=[3+?17]/4 or x2=[3-?17]/4
x1=[3+4.123]/4 or x2=[3-4.123]/4
x1=1.78 or x2=-0.28
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