Question

已知x1和x2是方程2x^2-mx+m=0的两个实数根，且x1^2+x2^2=3，求m的值

Answer 1

x1+x2=m/2 x1*x2=m/2
(x1+x2)*(x1+x2)-2* x1*x2 = x1^2+x2^2
m/2 *m/2 -2*m/2 =3
解得 m=6或-2
m=6时，方程的b*b-4*a*c=m*m-8m<0方程无解
所以m=-2

Answer 2

韦达定理：
x1+x2=m/2
x1*x2=m/2
2x1*x2=m
x1²+x2²=3
x1²+x2²+2x1x2=3+m
(x1+x2)²=3+m
(m/2)²=3+m
m²-4m-12=0
(m-6)(m+2)=0
m1=6;m2=-2.

Answer 3

x1+x2=m/2
x1x2=m/2
x1^2+x2^2=（x1+x2）²-2x1x2
=m²/4-m=3，
m²-4m-12=0
(m+2)(m-6)=0
m=-2,m=6
有因为Δ=m²-8m>=0 (x1,x2是实根)
m<=0或m>=8
所以
m=-2

Answer 4

解；由韦达定理
x1+x2=m/2 (1)
x1*x2=m/2 (2)
(1)²-（2）*2
x1²+x2²=m²/4-m=3
m²-4m-12=0
(m-6)(m+2)=0
m=6或m=-2
经检验，m=-2为所求