We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
64
3
avatar+1206 

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+101813 
+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+10413 
0

Find the smallest possible value of a.

laugh

 Jun 20, 2019

7 Online Users