Question

不解方程，求作一个新的一元二次方程，使它两根分别是5x^2+2x-3的两根的平方的负倒数

Answer 1

设5x^2+2x-3=0的两根为x1,x2,则
x1+x2=-2/5,x1x2=-3/5,
-1/x1^2-1/x2^2=-(x1^2+x2^2)/(x1x2)^2=-[(x1+x2)^2-2x1x2]/(x1x2)^2
=-(4/25+6/5)/(9/25)=-34/9,
(-1/x1^2)(-1/x2^2)=1/(x1x2)^2=25/9,
∴所求的新方程为x^2+(34/9)x+25/9=0,即9x^2+34x+25=0.

Answer 2

设所求方程为y²+by+c=0，两根为y1，y2
则y1+y2=-b，y1*y2=c
设方程5x²+2x-3=0两根为x1，x2
则x1+x2=-2/5，x1*x2=-3/5

因为y1=-1/x1²，y2=-1/x2²
所以y1+y2=-(1/x1²+1/x2²)
=-(x1²+x2²)/(x1*x2)²
=-[(x1+x2)²-2x1*x2]/(x1*x2)²
=-[(-2/5)²-2*(-3/5)]/(-3/5)²
=-34/9
b=34/9

y1*y2=(-1/x1²)*(-1/x2²)=1/(x1*x2)²=25/9
c=25/9

所以所求方程为：y²+34/9y+25/9=0
整理为：9y²+34y+25=0

Answer 3

假设x1,x2是5x^2+2x-3=0两根，则
x1+x2=-2/5,x1x2=-3/5,x1^2+x2^2=(x1+x2)^2-2x1x2=4/25+6/5=34/25
(-1/x1^2)+(-1/x2^2)=-(x1^2+x2^2)/(x1^2*x2^2)=-34/25/(x1x2)^2=-34/25*25/9=-34/9
-1/x1^2*(-1/x2^2)=1/(x1x2)^2=25/9
所以新方程为9x^2+34x+25=0