Question 763902
Let the roots be {{{a}}} and {{{a/2}}}.
If we knew the coefficients and {{{a}}} was a nice round number we may solve such an equation by factoring.
Even when the roots are ugly irrational numbers, a polynomial can be written in factored form if you know the zeros.
In this case, the factored equation would be
{{{2(x-a/2)(x-a)=0}}} or {{{(2x-a)(x-a)=0}}}
Re-multiplying, we get
{{{2x^2-3ax+a^2=0}}}
If that equation is equivalent to {{{2x^2-kx+64=0}}} (both with {{{2}}} for a leading coefficient),
then the other copefficientsa are also the same, so
{{{a^2=64}}} and {{{-3a = -k}}}-->{{{3a=k}}}
{{{system(a^2=64,3a=k)}}} --> {{{system(a=8,or,a=-8)}}} --> {{{system(k=24,or,k=-24)}}}