方程X^2+(2M+1)X+M^2-2=0的两根平方和等于11,求M.___are needed.A.other two B.two else C.two more D.the other two
问题描述：
方程X^2+(2M+1)X+M^2-2=0的两根平方和等于11,求M.
___are needed.A.other two B.two else C.two more D.the other two
答
由题意可知：X1^2+X2^2=11；
X1+X2=-(2M+1);
X1*X2=M^2-2;
X1^2+X2^2=（X1+X2)^2-2* X1*X2=（2M+1）^2-2*(M^2-2)=11,
解上述方程求得M=1或M=-3
答
x1+x2=-b/a=-（2m+1），x1x2=c/a=m²-2
x1²+x2²
=（x1+x2）²-2x1x2
=（2m+1）²-4（m²-2）
=4m²+4m+1-2m²+4
=2m²+4m+5
即2m²+4m+5=11
2m²+4m-6=0
（2m+6）（m-1）=0
2m+6=0，m-1=0
得m1=-3，m2=1
答
设两根分别为x1,x2,由韦达定理得
x1+x2=-(2m+1)
x1x2=m^2 -2
x1^2+x2^2=(x1+x2)^2-2x1x2
=[-(2m+1)]^2-2(m^2-2)
=4m^2+4m+1-2m^2+4
=2m^2+4m+5=11
2m^2+4m-6=0
m^2+2m-3=0
(m+3)(m-1)=0
m=-3或m=1
___are needed.A.other two B.two else C.two more D.the other two
选D