Question

初三数学第二题

Answer 1

设方程6x^2-3x-2=0的两根为x1,x2
(x-x1)(x-x2)=0 x^2-(x1+x2)x+x1x2=0
x1+x2=2 x1x2=-1/3
x1^2+x2^2=(x1+x2)^2-2x1x2=4+2/3=14/3
x1^2x2^2=(x1x2)^2=1/9
设一元二次方程为(x-x1^2)(x-x2^2)=0

x^2-(x1^2+x2^2)x+x1^2x2^2=0
x^2-14/3x+1/9=0

追答

设方程6x^2-3x-2=0的两根为x1,x2
(x-x1)(x-x2)=0 x^2-(x1+x2)x+x1x2=0
x1+x2=1/2 x1x2=-1/3
x1^2+x2^2=(x1+x2)^2-2x1x2=1/4+2/3=11/12
x1^2x2^2=(x1x2)^2=1/9
设一元二次方程为(x-x1^2)(x-x2^2)=0
x^2-(x1^2+x2^2)x+x1^2x2^2=0
x^2-11/12x+1/9=0
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