Question

设X1、X2是方程“X平方减X减4等于0的两根，求X1的三次方加5倍X2的平方加10的值。

Answer 1

X²-X-4=0的两根为X1和
X2
所以X1+
X2
=1
X1X2=-4
把X1代入得X1²-X1-4=0
两边同时乘以X1得
X1³-X1²-4X1=0
X1³=X1²+4X1
把
X2代入得
X2²-X2-4=0
X2²=X2+4
把上面变形代入下式：
X1³+5X2²+10
=X1²+4X1+X2²+4X2²+10
=X1²+X2²+4X1+4X2²+10
=（X1+X2）²-2X1X2+4X1+4(X2+4)+10
=(X1+X2）²-2X1X2+4X1+4X2+16+10
=39

Answer 2

解：X²-X-4=0的两根为X1和
X2
所以X1+
X2
=1
X1X2=-4
把X1代入得X1²-X1-4=0
两边同时乘以X1得
X1³-X1²-4X1=0
X1³=X1²+4X1
把
X2代入得
X2²-X2-4=0
X2²=X2+4
把上面变形代入下式：
X1³+5X2²+10
=X1²+4X1+X2²+4X2²+10
=X1²+X2²+4X1+4X2²+10
=（X1+X2）²-2X1X2+4X1+4(X2+4)+10
=(X1+X2）²-2X1X2+4X1+4X2+16+10
=39