Question 1065005
I agree that {{{3/4sqrt(2) >= k}}} is necessary
to get two real solutions, {{{d}}} and {{{e}}} ,
but we need more than that,
and trying to use the quadratic formula seems awkward in this case.
I would say that it is necessary and sufficient
to have two real solutions, {{{d}}} and {{{e}}} ,
such that {{{system(d>=0,e>=0,d^2+e^2=1)}}} .
For any quadratic equation, we know that
if {{{d}}} and {{{e}}} are solutions to {{{ax^2 +bx + c = 0}}} ,
{{{system(de=c/a,"and",d+e=-b/a))}}} .
In this case,
if {{{d}}} and {{{e}}} are solutions to {{{sqrt(2)x^2 - sqrt(3)x + k = 0}}} ,
{{{system(de=k/sqrt(2),"and",d+e=sqrt(3)/sqrt(2))}}} .
With {{{k>=0}}} , we would have {{{de=k/sqrt(2)>0}}} , meaning {{{d>=0}}} and {{{e>=0}}} .
Let us see if we can find a {{{k>=0}}} out of
{{{system(d^2+e^2=1,de=k/sqrt(2),d+e=sqrt(3)/sqrt(2))}}} .
{{{system(d^2+e^2=1,de=k/sqrt(2),d+e=sqrt(3)/sqrt(2))}}} --> {{{system(d^2+e^2=1,2de=2k/sqrt(2),(d+e)^2=3/2)}}} --> {{{system(d^2+e^2=1,2de=2k/sqrt(2),d^2+e^2+2de=3/2)}}} --> {{{system(d^2+e^2=1,2de=2k/sqrt(2),1+2k/sqrt(2)=3/2)}}} --> {{{system(d^2+e^2=1,2de=2k/sqrt(2),2k/sqrt(2)=1/2)}}} --> {{{system(d^2+e^2=1,2de=2k/sqrt(2),highlight(k=sqrt(2)/4))}}}
 
NOTE:
The solutions to {{{sqrt(2)x^2 - sqrt(3)x + k = 0}}} are
{{{x=(sqrt(3) +- 1)/(2sqrt(2))}}} ,
which correspond to 
{{{system(theta=pi/12,"or",theta=5pi/12)}}}