已知x1+x2是方程x²-2x+a=0的两个实数根,且x1+2 x2=3-根号2,①求x1,x2及a的值②求x1³-3x1³+2x1+x2的值

已知x1+x2是方程x²-2x+a=0的两个实数根,且x1+2 x2=3-根号2,①求x1,x2及a的值②求x1³-3x1³+2x1+x2的值
已知x1+x2是方程x²-2x+a=0的两个实数根,且x1+2 x2=3-根号2,
①求x1,x2及a的值
②求x1³-3x1³+2x1+x2的值

已知x1+x2是方程x²-2x+a=0的两个实数根,且x1+2 x2=3-根号2,①求x1,x2及a的值②求x1³-3x1³+2x1+x2的值
数学之美为您解答
根据韦达定理可知,X1+X2=2 X1*X2=a
x1+2 x2=3-√2
所以,X1=1+√2 X2=1-√2 a=-1
x1³-3x1³+2x1+x2
=X1(x1³-3X1+2)+X2
=X1(X1-1)(X1-2)+X2
=(1+√2)*√2*(√2-1)+1-√2
=1