一元二次方程
1、解:由韦达定理得:x1+x2=-4
x1三次方+14x2+50
=[x1(x1²+4x1+2)-4(x1²+4x1+2)+14x1+8]+14x2+50
=14(x1+x2)+58
=-56+58
=2
2、由韦达定理:x1+x2=1-2m
x1·x2=m²+1
(x1²-x2²)²=(x1²+x2²)²-4x1²x2²
=[(x1+x2)²-2x1x2]²-4(x1x2)²
=[(1-2m)²-2×(m²+1)]²-...全部
1、解:由韦达定理得:x1+x2=-4
x1三次方+14x2+50
=[x1(x1²+4x1+2)-4(x1²+4x1+2)+14x1+8]+14x2+50
=14(x1+x2)+58
=-56+58
=2
2、由韦达定理:x1+x2=1-2m
x1·x2=m²+1
(x1²-x2²)²=(x1²+x2²)²-4x1²x2²
=[(x1+x2)²-2x1x2]²-4(x1x2)²
=[(1-2m)²-2×(m²+1)]²-4×(m²+1)²
=-16m^3-4m²+8m-3
有:-16m^3+4m²+8m-3=0
16m^3-4m²-2m-6m+3=0
2m(2m-1)(m+4)-3(2m-1)=0
(2m-1)(2m²+8m-3)=0
m1=1/2,m2=-4+√22,x3=-4-√22
经过检验m=1/2符合题意。
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