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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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?

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  • $\begingroup$ 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

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