Question

设方程4x2－7x+3=0的两根为x1,x2，不解方程，求x1x22＋x1/2 的值:

x2是x的平方，x1x22是x1与x2的平方的积，x1/2 是x1的一半

Answer 1

X1+X2=7/4，X1*X2=3/4
原式=（X1*X2)*X2+X1*0.5
=0.75*X2+0.5*X1
=6/8*X2+4/8*X1
=5/8(X1+X2)+1/8（x2-x1)
=5/8*7/4{+-}1/8*根号下[(7/4)^2-4*3/4]
=35/32{+-}1/8*1/4
=9/8或17/16

Answer 2

X1+X2=7/4，X1*X2=3/4 x2-x1=根号下[(X1+X2)^2-4X1*X2]
x1x22＋x1/2=（X1*X2)*X2+X1*0.5=0.75*X2+0.5*X1=6/8*X2+4/8*X1=5/8(X1+X2)+1/8（x2-x1)=5/8*7/4{+-}1/8*根号下[(7/4)^2-4*3/4]=35/32{+-}1/8*1/4=9/8或17/16

Answer 3

求不出来？好好看看题目？