设6x^2+x-1=0两根为x1,x2,新方程的两根为y1,y2
则y1=x1+1,y2=x2+1
由韦达定理,得,
x1+x2=-1/6,x1*x2=-1/6
所以y1+y2
=(x1+1)+(x2+1)
=x1+x2+2
=(-1/6)+2
=11/6
y1*y2
=(x1+1)(x2+1)
=x1*x2+x1+x2+1
=(-1/6)+(-1/6)+1
=2/3
所以新方程为y²-11y/6+2/3=0,
即6y²-11y+4=0
追问题目要求是6x的三次方,那么有解吗?
追答方程6x^3+x-1=0只有一个根,题目有误!
这是图像,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b3119313b07eca8000550e62912397dda04483e6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)