Let the roots be p and q.

Then p+q = -k and pq=36

Solve the 2nd equation for p and substitute in the first and you'll get:

 and 

Pick any nonzero number for one root q, then the other root p will be 36/q,
and k will be .  Here are some possible values.  


if q = -10, then p = -18/5 and k = 68/5
if q = -9, then p = -4 and k = 13
if q = -8, then p = -9/2 and k = 25/2
if q = -7, then p = -36/7 and k = 85/7
if q = -6, then p = -6 and k = 12   
if q = -5, then p = -36/5 and k = 25/2
if q = -4, then p = -9 and k = 13
if q = -3, then p = -12 and k = 15
if q = -2, then p = -18 and k = 20
if q = -1, then p = -36 and k = 37
if q = 1, then p = 36 and k = -37
if q = 2, then p = 18 and k = -20
if q = 3, then p = 12 and k = -15
if q = 4, then p = 9 and k = -13
if q = 5, then p = 7.2 and k = 25/2
if q = 6, then p = 6 and k = -12
if q = 7, then p = 36/7 and k = -85/7
if q = 8, then p = 4.5 and k = -25/2
if q = 9, then p = 4 and k = -13
if q = 10, then p = 18.5 and k = -137/2

I lined through 2 of them because the roots are not different in those.

Edwin

Since the problem talks about different real roots of a quadratic equation, it means that its discriminant is positive.


The discriminant is  b^2 - 4ac = k^2 - 4*36 = k^2 - 144.


So, the discriminant must be positive

    k^2 - 144 > 0.



It implies

    k^2 > 144, 



which means that  EITHER k > 12  OR  k < - 12.



ANSWER.  The set of all possible values of k is  { k |  k > 12  or  k < - 12 }.


         It is the union of two infinite semi-intervals  (-oo,12) U (12,oo).