+1245 Let a be a positive real number such that all the roots of \(x^3 + ax^2 + ax + 1 = 0\) are real. Find the smallest possible value of a.
+129899 x^3 + ax^2 + ax + 1 = 0
First....notice that no matter what "a" might be, -1 is a root because
(-1)^3 + a(-1)^2 + a(-1) + 1 =
-1 + a - a + 1 = 0
Using synthetic division
-1 [ 1 a a 1 ]
-1 1 - a -1
_________________________
1 a - 1 1 0
So....the remaining polynomial is x^2 + (a - 1)x + 1
The discriminant of this must be ≥ 0 for there to be remaining real roots....so.....
(a - 1)^2 - 4(1)(1) ≥ 0
a^2 - 2a + 1 - 4 ≥ 0
a^2 - 2a - 3 ≥ 0
(a - 3) ( a + 1) ≥ 0
This will be true on these intervals (-inf, -1] and [3, inf)
So.....since a is positive......then its smallest value = 3
