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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.

 Jun 20, 2019
 #1
avatar+35 
+1

a = 4?

 Jun 20, 2019
 #2
avatar+111401 
+2

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

 

 

cool cool cool

 Jun 20, 2019
edited by CPhill  Jun 20, 2019
 #3
avatar+12271 
0

Find the smallest possible value of a.

laugh

 Jun 20, 2019

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