Question 927517
Find all real values of {{{b}}} such that the equation: 

{{{x^2 + bx + 6b = 0}}} only has integer roots.

Theorem:
Let  {{{f(x)=ax^2 + bx + c}}}, being a quadratic trinomial with integer coefficients {{{a}}}, {{{b}}}, {{{c}}}. Then, both roots or zeros of {{{f(x)}}} are integers; if, and only if,
(i)
The integer {{{D=b^2-4ac}}} is an integer or perfect square. 
and 
(ii) The leading coefficient {{{a}}} is a divisor of both {{{b}}} and {{{c}}}. 


in your case, The leading coefficient {{{a}}} is a divisor of both {{{b}}} and {{{c}}} because {{{a=1}}} 

but, we need {{{b}}}; so,

{{{x^2 + bx + 6b = 0 }}} if discriminant {{{D>0}}} (Positive Discriminant ), we will have Two Real Solutions

{{{D=b^2-4ac}}}

{{{b^2-4ac>0}}}

{{{b^2-4*1*6b>0}}} ........solve for {{{b}}}

{{{b^2-24b>0 }}}

{{{b^2>24b}}}

{{{b>24}}}

solution is {{{b>24}}}

let have first one greater then{{{24}}}:{{{ b=25}}}

{{{highlight(x^2 + 25x + 6*25 = 0)}}} ...only has integer roots

proof:

{{{x^2 + 25x + 150 = 0}}}

{{{x^2 + 15x +10x+ 150 = 0}}}

{{{(x^2 + 15x) +(10x+ 150 )= 0}}}

{{{x(x + 15) +10(x+ 15 )= 0}}}

{{{(x+10) (x+15) = 0}}}

{{{x = -15}}}  and {{{x=-10}}} ->solutions are integer roots


{{{ graph( 600, 600, -20, 5, -10, 170, x^2 + 25x + 150) }}}