關(guān)于x的方程x²-(k+1)x+k-1=0中,a=1 ,b=-(k+1) ,c=k-1
方程判別式△=b²-4ac
=[-(k+1)]²-4×1×(k-1)
=k²+2k+1-4k+4
=k²-2k+5
=(k-1)²+4
∵(k-1)²+4≥4
∴△﹥0,方程有兩個不相等的實數(shù)根.為什么（k-1）²＋4≧4你好：同學(xué)！∵(k-1)²≥0∴(k-1)²+4≥4