Question 891093
<pre>
{{{3x^2-(3k+2)x+18=0}}}

We make the left side into a "monic" polynomial.  A "monic" polynomial
is one with the coefficient of the term with the greatest degree as 1.
So we divide each coefficient by 3:

{{{(3/3)x^2-((3k+2)/3)x+(18/3)=(0/3)}}}

{{{x^2-((3k+2)/3)x+6=0}}}

In a "monic" polynomial of degree n, the coefficient of the term in
x<sup>n-1</sup> is -1 times the sum of the roots of the polynomial.

So the sum of the roots is {{{-1*expr(-(3k+2)/3)}}} = {{{(3k+2)/3}}}

Set this equal to 6:

{{{(3k+2)/3}}}{{{""=""}}}{{{6}}}

 {{{3k+2}}}{{{""=""}}}{{{18}}}

{{{3k}}}{{{""=""}}}{{{16}}}

{{{k}}}{{{""=""}}}{{{16/3}}}

Edwin</pre>