Question

已知方程2x²-mx-2m+1=0的两个实数根是平方和为4分之29,求m的值

Answer 1

x1+x2=m/2
x1x2=(-2m+1)/2
x1^2+x2^2=(x1+x2)^2-2x1x2
=m^2/4-2*(-2m+1)/2
=m^2/4+2m-1
=29/4
即
m^2+8m-4=29
m^2+8m-33=0
(m-3)(m+11)=0
m=3,m=-11

Answer 2

x1+x2=m／2, x1·x2=﹙1－2m﹚／2
x1²＋x2²=﹙x1＋x2﹚²－2x1·x2
m²／4－﹙1－2m﹚＝29／4
m²－4﹙1－2m﹚＝29
m²＋8m－4=29
m²＋8m－33=0
(m+11)(m-3)=0
m1=-11, m2=3
∵⊿≥0,m=-11不合题意
∴取m=3

Answer 3

x1+x2=-b/a=m/2
x1x2=c/a=-m+1/2
x1^2+x^2=(x1+x2)^2-2x1x2=m^2/4+2m-1=29/4
m^2+8m-33=0
(m+11)(m-3)=0
m=-11或3
检验b^2-4ac>0
m^2+16a-8>0
当m=-11，m^2+16a-8=-63<0，
m=3时，m^2+16a-8=49>0
所以m=3

Answer 4

设2根为a,b
a^2+b^2=29/4
(a+b)^2-2*a*b=29/4
(m/2)^2-2*(-(2*m-1)/2)=29/4
m^2+8*m-33=0
m=3或m=-11
当m=-11时
方程为
2x²+11x+23=0
因为11^2-4*2*23<0，方程无实根，
所以m=3