Question

已知方程x^2-7x+8=0的两根为x1,x2,求作一个新的一元二次方程，

（1）使它的两根为原方程两根倒数的相反数
（2）使它的一根为原方程两根和的倒数，另一根为原方程两根差的平方

Answer 1

x1`=-1/x1,x2`=-1/x2
x1+x2=7,x1x2=8
x1`+x2`=-1/x1-1/x2=-(1/x1+1/x2)=-(x1+x2)/x1x2=-(7)/8=-7/8
x1`x2`=1/x1*1/x2=1/x1x2=1/8
x^2+7/8x+1/8=0
8x^2+7x+1=0
(2)x1`=1/(x1+x2)=1/7
x2`=(x1-x2)^2=(x1+x2)^2-4x1x2=49-4*8=49-32=17
x1`+x2`=1/7+17=(1+7*17)/7=120/7
x1`x2`=1/7*17=17/7
x^2-120/7x+17/7=0
7x^2-120x+17=0

Answer 2

x1+x2=7,x1x2=8
设新方程为x²+ax+b=0
(1)
-a=-1/x1-1/x2=-(x2+x1)/x1x2=-7/8
b=(-1/x1)(-1/x2)=1/x1x2=1/8
x²-7x/8+1/8=0
8x²-7x+1=0
(2)
-a=(1/x1+x2)+(x1-x2)²=(1/x1+x2)+[(x1+x2)²-4x1x2]=(7²-4*8)+1/7=120/7
b=(1/x1+x2)*(x1-x2)²=(1/x1+x2)*[(x1+x2)²-4x1x2]=(7²-4*8)*1/7=17/7
x²-120x/7+17/7=0
7x²-120x+17=0