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

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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

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
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

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
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

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

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