Question

已知方程2x平方-4x+1=0的两根分别为x1,x1,求下列各式的值

Answer 1

由已知可推出 x1+x2=2 、 x1*x2=1/2

∴ 1) 1/x1+1/x2=(x1+x2)/(x1*x2)=2/(1/2)=2*2=4

2) (x1+1)(x2+1)=x1*x2+x1+x2+1=1/2+2+1=7/2

3) x2/x1+x1/x2=(x2^2+x1^2)/(x1*x2)=(x1^2+2x1*x2+x2^2-2x1*x2)/(x1*x2)

=(x1^2+2*x1*x2+x2^2)/(x1*x2)-（2x1*x2)/(x1*x2)

=(x1+x2))^2/(x1*x2)-2

=(2^2)/(1/2)-2

=8-2

=6

4) (x1-x2)^2=(x1+x2)^2-4x1*x2

=2^2-4*(1/2)

=4-2

=2