Question

已知X1,X2是方程2X^2-7X-4=0的两个根，不解方程求X1(2X1-7)+(X1-1)(X2-1),2X1^2+4X2^2-7X2

Answer 1

X1,X2是方程2X^2-7X-4=0的两个根
所以x1+x2=7/2 x1x2=-2 2x1^2-7x1-4=0 2X2^2-7x2-4=0
X1^2+X2^2=(x1+x2)^2-2x1x2=(7/2）^2+4=55/4
X1(2X1-7)+(X1-1)(X2-1)
=2x1^2-7x1+x1x2-x2-x1+1
=(2x1^2-7x1-4+4)+x1x2-(x1+x2)+1
=(2x1^2-7x1-4)+4+x1x2-(x1+x2)+1
=0+4+（-2）-7/2+1
=-1/2
2X1^2+4X2^2-7X2
=2X1^2+2X2^2+(2X2^2-7x2-4)+4
=2( X1^2+X2^2)+(2X2^2-7x2-4)+4
=2(55/4)+0+4
=63/2

Answer 2

解：由，题意得
X1+X2=7/2
X1乘以X2=-2
X1,X2满足方程，得
2X1^2-7X1-4=0
2X2^2-7X2-4=0
因此，
X1(2X1-7)+(X1-1)(X2-1)=2X1^2+X1X2-8X1-X2+1
= (2X1^2-7X1-4)-(X1+X2)+X1X2+5
=0-7/2+(-2)+5
=-1/2
2X1^2+4X2^2-7X2=( 2X1^2-7X1-4)+2( 2X2^2-7X2-4)+7(X1+X2)+12
=0+2乘以0+7乘以7/2+12
=73/2