Question

一元二次方程对任意实数k,方程（k+1）x²-3（k+m）x+4kn=0总有一根为1,求m,n的值,并解此方程.

Answer 1

x=1

k+1-3k-3m+4kn=0恒成立

(4n-2)k=3m-1

所以4n-2=3m-1=0

所以m=1/3,n=1/2

(k+1)x²-3(k+1/3)x+2k=0

则x1+x2=(3k+1)/(k+1)

x1x2=2k/(k+1)

x1=1

所以x2=2k/(k+1)