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\header{Math 3010}{HW 7}{YOUR NAME HERE}
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\begin{document}

\begin{questions}

  \question Omar Khayyam developed a technique which used the point of
  intersection of two conic sections in the plane to geometrically
  estimate solutions to cubic polynomial equations. In our class
  discussion, I mentioned that the following equation is actually the
  equation of a circle in disguise:
  \[
  \dfrac{q-x}{y} = \dfrac{y}{x}.
  \]
  Show that this equation is equivalent to the equation of a circle
  with center located at $(q/2,0)$.

  \begin{solution}
    Your answer here...
  \end{solution}

  \question Solve the following cubic equation in $y$ via Tartaglia's
  method.
  \[
  y^3 - 3y^2 - y + 3 = 0
  \]

  First, \emph{reduce} it to a cubic equation in $x$ with no
  quadratic term, i.e. an equation of the form $x^3 + px + q = 0$.

  (Hint 1: Set $y=x+k$, and then choose $k$ such that the coefficient of
  the quadratic term is 0.)

  Once you know $p$ and $q$ use Cardano's formula below to solve for
  $x$ and finally use your values for $x$ and $k$ to solve for the
  root $y = x+k$.

  (Hint 2: Cardano's formula will give you an expression involving
  complex numbers, but this expression simplifies to an integer.)

  \begin{equation}
    x = \sqrt[3]{-\dfrac{q}{2} + 
      \sqrt{\left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3}} +
    \sqrt[3]{-\dfrac{q}{2} -
      \sqrt{\left(\frac{q}{2}\right)^2 + \left(\frac{p}{3}\right)^3}}
    \tag{Cardano's Formula}
  \end{equation}

  Show your work to receive full credit.

  \begin{solution}
    Your answer here...
  \end{solution}

\end{questions}
\end{document}
