Question

使方程x^2-ax+9a=0 的根都为整数的实数a的所有徝之和为

Answer 1

设根是m和n
则m+n=a
mn=9a
所以mn=9m+9n
(n-9)m=9n
m=9n/(n-9)=(9n-81+81)/(n-9)=9(n-9)/(n-9)+81/(n-9)=9+81/(n-9)
m是整数
所以n-9是81的约数
所以n-9=1,-1,3,-3,9,-9,27,-27,81,-81
n=10,7,12,6,18,0,36,-18,90,-72
则相应的m=90,-63,36,-18,18,0,12,6,10,8
a=m+n
所以a=100,-54,48,-12,36,0,48,-12,100,-64
所以a有7个值
a=100,-54,48,-12,36,0,-64
所以a的所有徝之和=100-54+48-12+36-64=54