Showing that $3x^2−x+k=0$ has roots $\frac{p}4$ and $p+1$ for unique values of $k$ and $p$ How do I continue the question with algebraic methods:

The two solutions of the quadratic equation $3x^2−x+k=0$ are $\dfrac{p}{4}$ and $p+1$. Determine the values of $k$ and $p$.

I asked this question roughly a week ago and a user by the name of Crease said I could substitute $x$ into the equation with the above values resulting in:
$$3\left(\dfrac{p}{4}\right)^2−\dfrac{p}{4}+k=0 \qquad
\text{and}\qquad 3(p+1)^2−(p+1)+p=0$$
I continued this and got two final values for $p$ (I only want one).
Crease also said I could simply use Vieta's relations to solve it, and so I did however I showed my teacher this, and he said that I wouldn't be able to use Vieta's relations on the final assessment.

How do I go about solving this problem through algebraic methods without receiving two values for $p$ or $k$?

Thanks in advance for anyone that wishes to help!
Edit: Original question is Finding the x values of a quadratic that has many different variables
 A: You can always derive Vieta's on the spot
$$\left( x-\frac{p}{4}\right)(x-(p+1))=0$$
$$x^2-\left( \frac{p}4+p+1\right)x+(p+1)\left( \frac{p}4\right)=0$$
$$x^2-\frac{x}{3}+\frac{k}{3}=0$$
By comparing coefficient, we have
$$\frac{p}{4}+p+1=\frac13$$
$$\frac{5p}{4}=-\frac23$$
$$p=-\frac{8}{15}$$
Now that we have the values of $p$, can you solve for $k$?

$$3\left( \frac{p}4\right)^2-\frac{p}4 + k = 0$$
$$3(p+1)^2 - (p+1)+k=0$$
Subtracting them:
$$3\left( \frac{p}{4}+p+1\right)\left( \frac{p}{4}-p-1\right)-\left( \frac{p}{4}-p-1\right)=0$$
$$\left( \frac{p}{4}+p+1-\frac13\right)\left( \frac{p}{4}-p-1\right)=0$$
$$\left( \frac{5p}{4}+\frac23\right)\left( \frac{-3p}{4}-1\right)=0$$
If $p=-\frac43$, then we have $p+1=\frac{p}4=-\frac13$ and the polynomial should be $$3\left(x+\frac13\right)^2=3x^2+2x+\frac13$$
which is not of the form that is stated.
A: A quadratic $ax^2+bx+c$ with roots $x_1,x_2$ factors as $a(x-x_1)(x-x_2)$. In this case:
$$
\require{cancel}
\begin{align}
& 3x^2 - x + k = 3\left(x-\frac{p}{4}\right)\left(x - (p+1)\right)
\\ \iff\;\;\;\; &12x^2-4x+4k = 3\left(4x-p\right)\left(x-(p+1)\right)
\\ \iff\;\;\;\; &\cancel{12x^2}-4x+4k = \cancel{12x^2} - (15p+12)x +p(p+1)
\\ \iff\;\;\;\; &
\begin{cases}
4 &= 15p + 12
\\ 4k &= p(p+1)
\end{cases}
\end{align}
$$
The first equation gives $p=-\frac{8}{15}$, then the second one gives $k$.

[ EDIT ]   The reason you got an extraneous value for $p$ is that, by doing the direct substitutions in the equation, you lost the condition that $p/4$ and $p+1$ must be the two roots of the equation, not the same root counted twice. The extra value $p=-4/3$ actually solves the different question "when is $p/4$ a root and at the same time $p/4=p+1$".
A: $3x^2-x+k=0$ has roots $\frac{p}{4}$ and $p+1$ (as long as I didn't misunderstood).
So $$ p(\frac{p}{4})= 3(\frac{p}{4})^2-\frac{p}{4}+k=0$$
Solving this we get $ k=\frac{4p-3p^2}{16} $
Now $$ p(p+1)=3(p+1)^2-(p+1)+k=0 $$
Solving this we get $ -(3p^2+5p+2)=k $
So we get  $ \frac{4p-3p^2}{16}=-(3p^2+5p+2) $
Solving this you get $p=\frac{-4}{3}, p=\frac{-8}{15}$
Thus we get the roots if $p=\frac{-4}{3}$ will be $\frac{-1}{3}$ and if $p=\frac{-8}{15}$ then roots are $\frac{-2}{15} and \frac{7}{15}$
Now the starting polynomial can be written $x^2-\frac{x}{3}+\frac{k}{3} = x^2-(\frac{p}{4}+p+1)+(\frac{p}{4}(p+1))$
Thus we can say $p=\frac{-8}{15}$ because $p=\frac{-4}{3}$ doesn't give the roots that satisfy the equation. If $p=\frac{-8}{15}$ then can you find k?
A: You can try plotting the graph of $y=3x^2-x+k$. It will be an upward opening parabola with points of intersection with x-axis as $(\frac{p}{4},0)$ and $(p+1,0)$.
So, the axis of the parabola will be the line $x=\frac{\frac{p}{4}+p+1}{2} \Rightarrow x=\frac{5p}{8}+\frac{1}{2}$.
$\Rightarrow$ The vertex of the parabola will have it's x coordinate $\frac{5p}{8}+\frac{1}{2}$.
Also, the vertex of a parabola $y=ax^2+bx+c$ has the coordinates $(\frac{-b}{2a},\frac{4ac-b^2}{4a})$, so we can apply this to solve for $p$.
$$\frac{5p}{8}+\frac{1}{2}=-\frac{-1}{2\times3}=\frac{1}{6}$$
$$\Rightarrow p=\frac{-8}{15}$$
Now you can substitute the value of either $\frac{p}{4}$ or $p+1$ in the given equation and solve for $k$.
