若方程x∧2-2(m+1)x+3m∧2-4mn+4n∧2+2=0有实根,那么实数m,n的值分别是多少
问题描述:
若方程x∧2-2(m+1)x+3m∧2-4mn+4n∧2+2=0有实根,那么实数m,n的值分别是多少
答
若方程x∧2-2(m+1)x+3m∧2-4mn+4n∧2+2=0有实根,则有:Δ≥0即:4(m+1)²-4(3m²-4mn+4n²+2)≥0(m+1)²-(3m²-4mn+4n²+2)≥0m²+2m+1-3m²-4mn-4n²-2≥0-m²+2m-1-m...