The question was as follows:

The equations $x^3+Ax+10=0$ and $x^3+Bx^2+50=0$ have two roots in common. Compute the product of these common roots.

Because $x^3+Ax+10=0$ and $x^3+Bx^2+50=0$ it means that $x^3+Ax+10=x^3+Bx^2+50$

Take $x^3+Ax+10=x^3+Bx^2+50$ and remove $x^3$ from both sides, you get $Ax+10=Bx^2+50$ or $Bx^2-Ax+40=0$

By the quadratic equation, we get $\frac {A \pm \sqrt {(-A)^2 - 4*40B}}{2B}=\frac {A \pm \sqrt {A^2 - 160B}}{2B}$

This gives us two answers: $\frac {A + \sqrt {A^2 - 160B}}{2B}$ and $\frac {A - \sqrt {A^2 - 160B}}{2B}$

$\frac {A + \sqrt {A^2 - 160B}}{2B} * \frac {A - \sqrt {A^2 - 160B}}{2B}=\frac {A^2 - {A^2 - 160B}}{4B^2}$

This simplifies as $\frac {160B}{4B^2}=\frac{40}{B}$

$\frac{40}{B}$ is an answer, but in the solutions, they expected an integer answer. Where did I go wrong?

  • $\begingroup$ They were expecting you to find the values of A and B and then use those to find the roots. $\endgroup$ – Aidan F. Pierce May 25 '14 at 19:37
  • $\begingroup$ You don't need to solve the quadratic to get $\frac {40}B$ - you ought to be able to read that off. $\endgroup$ – Mark Bennet May 25 '14 at 19:47
  • $\begingroup$ Essentially, setting the equations equal to each other and solving ignores the information that the numbers in question are $\textit{roots}$. $\endgroup$ – Peter Woolfitt May 25 '14 at 19:52
  • $\begingroup$ Then how do I set it up to find the actual values? $\endgroup$ – Asimov May 25 '14 at 20:07
  • $\begingroup$ The solutions to the quadratic $\frac {A \pm \sqrt {A^2 - 160B}}{2B}$ represent where the cubics intersect. Plugging this solution back into each cubic equation to find the constraints on A and B. $\endgroup$ – David H May 25 '14 at 20:37

Let $a,b$ be the roots.

Then $a,b$ are roots of $$x^3+Ax+10=0$$

The sum of the three roots of this polynomial is negative the coefficient of $x^2$, thus $0$. It follows that the third root is $-(a+b)$.

As the product of the three roots is $-10$ we get $$ab(a+b)=10$$

Now let $c$ be the third root of $$x^3+Bx^2+50=0$$ Then $$ab+ac+bc =0$$ or $$ab+c(a+b)=0$$ and $$abc=-50$$

Replacing $a+b=\frac{10}{ab}$ we get $$(ab)^2+10c =0$$ $$abc=-50$$

Multiply the first of these two equations by $ab$ and you are done.

| cite | improve this answer | |
  • $\begingroup$ Can you confirm what the final value is? OP claims it needs to be an integer, though it seems like you have the same answer as I do. $\endgroup$ – Calvin Lin May 26 '14 at 14:19
  • $\begingroup$ @CalvinLin I get $(ab)^3=500$, so not an integer... $\endgroup$ – N. S. May 26 '14 at 14:21

Hint: The common roots must be both roots of

$$- (x^3 + Ax +10 ) + (x^3 + Bx^2 + 50) = Bx^2 - Ax + 40 $$

Let this quadratic polynomial be denoted by $f(x)$.

Hint: We have

$$ f(x) ( \frac{1}{B} x + \frac{5}{4} ) = x^3 + Bx^2 + 50. $$

This gives $B^2 = 4A$ and $160=5AB$, so $5B^3 = 640 $. This gives $B = 4 \sqrt[3]{2} $, $ A = 4\sqrt[3]{4}$.

This does not give me an integer answer for $ \frac{40}{B} = 5 \sqrt[3]{4}$, so perhaps they had an error?

| cite | improve this answer | |
  • $\begingroup$ Thanks a ton. I got the first part but I didnt realize that i should have multiplied it by $(\frac{1}{B}+\frac{1}{4})$ $\endgroup$ – Asimov May 25 '14 at 20:59
  • $\begingroup$ Note that you need to do a bit more work, to find the value of B. $\endgroup$ – Calvin Lin May 25 '14 at 21:01
  • $\begingroup$ Yes, but that should be simple once the single formula is derived. $\endgroup$ – Asimov May 25 '14 at 21:02
  • $\begingroup$ @Asimov Note that I didn't get an integer answer. $\endgroup$ – Calvin Lin May 25 '14 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.