Question

设x1、x2是方程是2x^2+4x-3=0的两个根,利用根与系数的关系,求下列各式的值:

Answer 1

(x1+1)(x2+1)

= x1x2 + x1 + x2 + 1

= (-3/2) + (-4/2) + 1

= -5/2

x1^2x2+x1x2^2

= (x1x2)(x1+x2)

= (-3/2)(-4/2)

= 3

x2/x1+x1/x2

= (x1^2+x2^2)/(x1x2)

= (x1^2+2x1x2+x2^2-2x1x2)/(-3/2)

= ((-4/2)^2-2(-3/2))/(-3/2)

= -14/3

(x1-x2)^2

= (x1+x2)^2 - 4x1x2

= (-4/2)^2 - 4(-3/2)

= 4 + 6

= 10