根据韦达定理,x1+x2=(k+1)/2,x1*x2=(k+3)/2
由两根之差为1,即|x1-x2|=1
得(x1-x2)^2=(x1+x2)^2-4x1x2=[(k+1)/2]^2-4(k+3)/2=1化简为:
k^2-6k-27=0 因式分解(k-9)(k+3)=0 解得k=9或者-3
简单数学题:若方程2X^2-(k+1)x+k+3=0上的两根之差为1,则K=—————
简单数学题:若方程2X^2-(k+1)x+k+3=0上的两根之差为1,则K=—————
数学人气:201 ℃时间:2019-08-22 09:58:21
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