Showing that $3x^2−x+k=0$ has roots $\frac{p}4$ and $p+1$ for unique values of $k$ and $p$

Question

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

Answer 1

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 $p/4=p+1$".

Answer 2

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.

Answer 3

$3x^2-x+k=0$ 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?

Answer 4

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 $k$.