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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

Blue
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Rodney Millan
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4 Answers4

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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.

Siong Thye Goh
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  • Yes, that uses Vieta's relations, I am looking for the way you would do it without the use of Vieta's relations, would anybody be able help? – Rodney Millan Jun 04 '21 at 05:26
  • @RodneyMillan No, this answer does not use Vieta's relations. If you remove the comment in the first line ("derive Vieta's on the spot"), which is not part of the actual proof, you won't find Vieta's relations being used anywhere along the way. – dxiv Jun 04 '21 at 06:40
  • Sorry @dvix, this comment was made before Siong added that into the answer, I am currently trying to wrap my head around the answer without Vieta's relations – Rodney Millan Jun 04 '21 at 06:42
  • @RodneyMillan Guess I didn't follow the whole history. Anyway, in the current form it's not using Vieta's. – dxiv Jun 04 '21 at 06:44
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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 $p/4=p+1$".

dxiv
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    Thank you dxiv, I actually managed to wrap my head around this method – Rodney Millan Jun 04 '21 at 06:53
  • @RodneyMillan It's worth keeping in mind that your way of solving it is also correct, but you must realize that the solutions have to be verified, because the possibility exists that extraneous solutions were introduced. (Actually, this may well be the point of the exercise, and the reason your teacher insisted on not using Vieta's). Here is another example of a similar problem, and a related post of mine about it. – dxiv Jun 04 '21 at 07:10
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$3x^2-x+k=0$ 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?

  • Does this answer your question or you have any doubts? –  Jun 04 '21 at 05:38
  • This is the answer that I got when I tried answering the question with algebraic methods, my only question is why doesn't p = -4/3 give roots that satisfy the equation? How do you prove that? – Rodney Millan Jun 04 '21 at 05:42
  • $p=-4/3$ gives roots $-1/3,-1/3$ However as you can read "Now the starting polynomial can be written $x^2-\frac{x}{3}+\frac{k}{3} = x^2-(\frac{p}{4}+p+1)+(\frac{p}{4}(p+1))$" p/4+p+1 will result in 1/3 but the roots -1/3+(-1/3) doesn't give 1/3 so p=-4/3 is eliminated. Any more doubts? –  Jun 04 '21 at 06:05
  • Sorry didn't use MathJax because in a hurry won't be active for some time but you can ask question if you have any doubt –  Jun 04 '21 at 06:06
  • Sorry didn't use MathJax because in a hurry won't be active for some time but you can ask question if you have any doubt –  Jun 04 '21 at 06:06
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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 $k$.

RiverX15
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