Cho `x^2+3x-2=0`
`=>x^2+2.x. 3/2+9/4-17/4=0`
`=>(x+3/2)^2=17/4`
`=>(x+3/2)^2=(\sqrt{17}/2)^2` hoặc `(x+3/2)^2=([-\sqrt{17}]/2)^2`
`@TH1: x+3/2=\sqrt{17}/2=>x=[\sqrt{17}-3]/2`
`@TH2: x+3/2=[-\sqrt{17}]/2=>x=[-\sqrt{17}-3]/2`
Vây nghiệm của đa thức là `x=[\sqrt{17}-3]/2` hoặc `x=[-\sqrt{17}-3]/2`
`x^2 + 3x-2=0`
`=> x^2 + 3/2x + 3/2x + 9/4 - 17/4 = 0`
`=> x( x + 3/2 ) + 3/2( x + 3/2 ) = 17/4`
`=> ( x + 3/2 )^2 = 17/4 =` `(` `\sqrt{17}/2 )^2 = -\sqrt{17}/2 )^2`
`=> x = (\sqrt{17} - 3)/2` hoặc `x = (-\sqrt{17} - 3)/2`