解方程:x^2+2x-根号(2x^2+6x+2)=3-x
3个回答

移项,合并同类项,原方程化为:x2+3x-3-Genhao(2x^2+6x+2)=0.令y=Genhao(2x^2+6x+2),则(y/2)^2=x^2+3x+1,原方程化为:(y/2)^2-y-4=0,即

y^2-4y-16=0,配方,得到(y-2)^2=20.y=2+genhao20,另外一个根是负数,舍掉.然后代如y=Genhao(2x^2+6x+2),得到:2x^2+6x+2=4+4genhao20+20