Question

已知方程3x²-4x=-1的两根是x1 x2，不解方程，求： 1.x2/x1 + x1/x2 2.（x1 - 2）（x2 - 2）

Answer 1

解：方程3x²-4x=-1可化为：3x²-4x+1=0
由根与系数的关系，有
x1+x2=4/3, x1x2=1/3
∴x2/x1 + x1/x2
=(x1²+x2²)/(x1x2)
=[(x1+x2)²－2x1x2]/(x1x2)
=[(4/3)²－2×(1/3)]/(1/3)
=[(16/9)-(2/3)]/(1/3)
=(10/9)／(1/3)
=10/3

(x1-2)(x2-2)
=x1x2-2x1-2x2+4
=x1x2-2(x1+x2)+4
=(1/3)-2×(4/3)+4
=(1/3)-(8/3)+4
=(-7/3)+4
=5/3.

Answer 2

3x^2-4x=-1
3x^2-4x+1=0
x1+x2=-b/a=4/3
x1*x2=c/a=1/3

x2/x1+x1/x2
=(x1^2+x2^2)/(x1*x2)
=[(x1+x2)^2-2x1*x2]/(x1*x2)
=(16/9-2/3)/(1/3)
=10/9÷1/3
=10/3

(x1-2)(x2-2)
=x1*x2-2(x1+x2)+4
=1/3-8/3+4
=5/3

Answer 3

x1+x2=4/3
x1x2=1/3
所以
1.x2/x1 + x1/x2
=(x1²+x2²)/x1x2
=[(x1+x2)²-2x1x2]/x1x2
=(16/9-2/3)/(1/3)
=10/3
2.（x1 - 2）（x2 - 2）
=x1x2-2(x1+x2)+4
=1/3-8/3+4
=-7/3+12/3
=5/3