Question 759364
{{{3x^2-x+k=0}}}
and {{{p/4}}} and {{{p+1}}} are the two solutions; and you are to find p and k.


Use the general solution for a quadratic equation, and make use of the discriminant.  The value of discriminant must be positive.  Those are my suggestions, although I have not yet actually tried to solve this problem.


{{{x=(1+- sqrt(1-4*3*k))/(2*3)}}} and you MUST have {{{(1-4*3*k)>0}}}


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based on that, {{{1>12k}}}
{{{1/12>k}}}
{{{k<1/12}}}


It makes me think the problem is open-ended.  Maybe use k=-2 or k=-3?.  You'll still have two solutions for x.

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Can you take it from here?