Question

已知，α，β为方程X²+4x+2=0的两个实数根。求α³+14β+50的值。

谢谢了！

Answer 1

α，β为方程X²+4x+2=0的两个实数根,
α+β=-4.,α²=-4α-2
α³+14β+50
=α(-4α-2)+14β+50
=-4α²-2α+14β+50
=-4(-4α-2)-2α+14β+50
=14α+14β+58
=14(α+β)+58
=14×（-4）+58=2

Answer 2

α，β为方程X²+4x+2=0的两个实数根,
所以 α+β=-4.,α²+4α+2=0
所以 α²=-4α-2

α³+14β+5
=α×α²+14β+50
=α(-4α-2)+14β+50
=-4α²-2α+14β+50
=-4(-4α-2)-2α+14β+50
=16α+8-2α+14β+50

=14α+14β+58
=14(α+β)+58
=14×（-4）+58
=2

Answer 3

α+β=-4.,α²=-4α-2
α³+14β+50
=α(-4α-2)+14β+50
=-4α²-2α+14β+50
=-4(-4α-2)-2α+14β+50
=14α+14β+58
=14(α+β)+58
=14×（-4）+58=2