Question

已知关于x的一元二次方程x^2+2(m+1)x+(3m^2+4mn+4n^2)=0有实根，则3m^2+2n^3=? 要有详细步骤 你的不对

Answer 1

已知关于x的一元二次方程x^2+2(m+1)x+(3m^2+4mn+4n^2)=0有实根
因为有实根，所以△≥0
因为△=b^2-4ac
=[2(m+1)]^2-4*（3m^2+4mn+4n^2+2)
=4m^2+8m+4-12m^2-16mn-16n^2-8
=-8m^2-16mn-16n^2+8m-4≥0
得2m^2+4mn+4n^2-2m+1≤0
既(m^2+4mn+4n^2)+(m^2-2m+1)≤0
(m+2n)^2+(m-1)^2≤0

即 (m+2n)=0 (m-1)=0

得m=1,n=-1/2

则3m^2+2n^3=3*1^2+2*(-1/2)^3=3-2*1/8=3-1/4=2.75

Answer 2

解：∵关于X的一元二次方程x2+2(m+1)x+（3m2+4mn+4n2+2）=0有实根∴△=[2(m+1)]^2-4(3m2+4mn+4n2+2）=-8m^2+8m-16mn-16n^2-4=-4m^2+8m-4-4m^2-16mn-16n^2=-4(m-1)^2-4(m+2n)^2≥0∵(m-1)^2≥0(m+2n)^2≥0∴-4(m-1)^2-4(m+2n)^2≤0故-4(m-1)^2-4(m+2n)^2=0即4(m-1)^2+4(m+2n)^2=0∴4(m-1)^2=0 4(m+2n)^2=0∴m=1 n=-m/2=-1/2故3m2+2n3 =3*1^2+2*(-1/2)^3=11/4