Skip to main content Contents Index Prev Up Next \(\newcommand{\identity}{\mathrm{id}}
\newcommand{\notdivide}{\nmid}
\newcommand{\notsubset}{\not\subset}
\newcommand{\lcm}{\operatorname{lcm}}
\newcommand{\gf}{\operatorname{GF}}
\newcommand{\inn}{\operatorname{Inn}}
\newcommand{\aut}{\operatorname{Aut}}
\newcommand{\Hom}{\operatorname{Hom}}
\newcommand{\cis}{\operatorname{cis}}
\newcommand{\chr}{\operatorname{char}}
\newcommand{\Null}{\operatorname{Null}}
\newcommand{\transpose}{\text{t}}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Exercises 17.6 Additional Exercises: Solving the Cubic and Quartic Equations
1.
Complete the square to solve the general quadratic equation
\begin{equation*}
ax^2 + bx + c = 0
\end{equation*}
to obtain
\begin{equation*}
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\text{.}
\end{equation*}
The discriminant of the quadratic equation \(\Delta = b^2 - 4ac\) determines the nature of the solutions of the equation. If \(\Delta \gt 0\text{,}\) the equation has two distinct real solutions. If \(\Delta = 0\text{,}\) the equation has a single repeated real root. If \(\Delta \lt 0\text{,}\) there are two distinct imaginary solutions.
2.
Show that any cubic equation of the form
\begin{equation*}
x^3 + bx^2 + cx + d = 0
\end{equation*}
can be reduced to the form \(y^3 + py + q = 0\) by making the substitution \(x = y - b/3\text{.}\)
3.
Prove that the cube roots of 1 are given by
\begin{align*}
\omega & = \frac{-1+ i \sqrt{3}}{2}\\
\omega^2 & = \frac{-1- i \sqrt{3}}{2}\\
\omega^3 & = 1\text{.}
\end{align*}
4.
Make the substitution
\begin{equation*}
y = z - \frac{p}{3 z}
\end{equation*}
for \(y\) in the equation \(y^3 + py + q = 0\) and obtain two solutions \(A\) and \(B\) for \(z^3\text{.}\)
5. Show that the product of the solutions obtained in (4) is \(-p^3/27\text{,}\) deducing that \(\sqrt[3]{A B} = -p/3\text{.}\)
6.
Prove that the possible solutions for \(z\) in (4) are given by
\begin{equation*}
\sqrt[3]{A}, \quad \omega \sqrt[3]{A}, \quad \omega^2 \sqrt[3]{A}, \quad \sqrt[3]{B}, \quad \omega \sqrt[3]{B}, \quad \omega^2 \sqrt[3]{B}
\end{equation*}
and use this result to show that the three possible solutions for \(y\) are
\begin{equation*}
\omega^i \sqrt[3]{-\frac{q}{2}+ \sqrt{\ \frac{p^3}{27} + \frac{q^2}{4}} } + \omega^{2i} \sqrt[3]{-\frac{q}{2}- \sqrt{\ \frac{p^3}{27} + \frac{q^2}{4}} }\text{,}
\end{equation*}
where \(i = 0, 1, 2\text{.}\)
7.
The discriminant of the cubic equation is
\begin{equation*}
\Delta = \frac{p^3}{27} + \frac{q^2}{4}\text{.}
\end{equation*}
Show that \(y^3 + py + q=0\)
has three real roots, at least two of which are equal, if \(\Delta = 0\text{.}\)
has one real root and two conjugate imaginary roots if \(\Delta \gt 0\text{.}\)
has three distinct real roots if \(\Delta \lt 0\text{.}\)
8.
Solve the following cubic equations.
\(\displaystyle x^3 - 4x^2 + 11 x + 30 = 0\)
\(\displaystyle x^3 - 3x +5 = 0\)
\(\displaystyle x^3 - 3x +2 = 0\)
\(\displaystyle x^3 + x + 3 = 0\)
9.
Show that the general quartic equation
\begin{equation*}
x^4 + ax^3 + bx^2 + cx + d = 0
\end{equation*}
can be reduced to
\begin{equation*}
y^4 + py^2 + qy + r = 0
\end{equation*}
by using the substitution \(x = y - a/4\text{.}\)
10.
Show that
\begin{equation*}
\left( y^2 + \frac{1}{2} z \right)^2 = (z - p)y^2 - qy + \left( \frac{1}{4} z^2 - r \right)\text{.}
\end{equation*}
11.
Show that the right-hand side of
Exercise 17.6.10 can be put in the form
\((my + k)^2\) if and only if
\begin{equation*}
q^2 - 4(z - p)\left( \frac{1}{4} z^2 - r \right) = 0\text{.}
\end{equation*}
12.
\begin{equation*}
z^3 - pz^2 - 4rz + (4pr - q^2) = 0\text{.}
\end{equation*}
Solving the resolvent cubic equation, put the equation found in
Exercise 17.6.10 in the form
\begin{equation*}
\left( y^2 + \frac{1}{2} z \right)^2 = (my + k)^2
\end{equation*}
to obtain the solution of the quartic equation.
13.
Use this method to solve the following quartic equations.
\(\displaystyle x^4 - x^2 - 3x + 2 = 0\)
\(\displaystyle x^4 + x^3 - 7 x^2 - x + 6 = 0\)
\(\displaystyle x^4 -2 x^2 + 4 x -3 = 0\)
\(\displaystyle x^4 - 4 x^3 + 3x^2 - 5x +2 = 0\)
