Question

已知x1x2是方程2x^2-3x-1=0的两个根求下列各式的值(1)(x1-x2)^2 (2)x1^2+3x1x2+x2^2

Answer 1

已知x1x2是方程2x^2-3x-1=0的两个根求下列各式的值(1)(x1-x2)^2 (2)x1^2+3x1x2+x2^2
解：因为x1x2是方程2x^2-3x-1=0的两个根
由韦达定理得：
x1+x2=3/2
x1x2=-1/2

(1) (x1-x2)²=(x1+x2)²-4x1x2
=(3/2)²-4*(-1/2)
=9/4+2
=17/4

(2) x1²+3x1x2+x2²
=(x1+x2)²+x1x2
=9/4-1/2
=7/4

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Answer 2

x1x2是方程2x^2-3x-1=0的两个根
由韦达定理得：
x1+x2=3/2
x1x2=-1/2
(1) (x1-x2)²=(x1+x2)²-4x1x2
=(3/2)²-4*(-1/2)
=9/4+2
=17/4
(2) x1²+3x1x2+x2²
=(x1+x2)²+x1x2
=9/4-1/2
=7/4

Answer 3

x1x2是方程2x²-3x-1=0

x1+x2=3/2,x1x2=-1/2(韦达定理)
(x1-x2)²

=(x1+x2)²-4x1x2
=13/4
(2)x1²+3x1x2+x2²

=(x1+x2)²+x1x2
=7/4