Since there is no constant term, one root is 0.
x(x-r)(x-s) = 0
case 1: |x^2+ax| >= 0
x^3 + x^2 + ax = 0
x^3 + x^2 + ax = x^3 - (r+s)x^2 + (rs)x
r+s = -1
rs = a
a/s + s + 1 = 0
s^2 + s + a = 0
r,s = ( -1 ± √(1-4a) )/2 a < 1/4
case 2: |x^2+ax| < 0
x^3 - x^2 - ax = 0
x^3 - x^2 - ax = x^3 - (r+s)x^2 + (rs)x
r+s = 1
rs = -a
-a/s + s - 1 = 0
s^2 - s - a = 0
r,s = ( 1 ± √(1+4a) )/2 a > -1/4
-1/4 < a < 1/4
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Posted by Larry
on 2024-06-26 18:32:14 |
solution
