Skip to main content
\(\newcommand{\identity}{\mathrm{id}} \newcommand{\notdivide}{\nmid} \newcommand{\notsubset}{\not\subset} \newcommand{\lcm}{\operatorname{lcm}} \newcommand{\gf}{\operatorname{GF}} \newcommand{\inn}{\operatorname{Inn}} \newcommand{\aut}{\operatorname{Aut}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\cis}{\operatorname{cis}} \newcommand{\chr}{\operatorname{char}} \newcommand{\Null}{\operatorname{Null}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

Section17.5Additional Exercises: Solving the Cubic and Quartic Equations

1

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}. \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. \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}} }, \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}. \end{equation*}

Show that \(y^3 + py + q=0\)

  1. has three real roots, at least two of which are equal, if \(\Delta = 0\text{.}\)

  2. has one real root and two conjugate imaginary roots if \(\Delta \gt 0\text{.}\)

  3. has three distinct real roots if \(\Delta \lt 0\text{.}\)

8

Solve the following cubic equations.

  1. \(x^3 - 4x^2 + 11 x + 30 = 0\)

  2. \(x^3 - 3x +5 = 0\)

  3. \(x^3 - 3x +2 = 0\)

  4. \(x^3 + x + 3 = 0\)

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). \end{equation*}
11

Show that the right-hand side of Exercise 17.5.10 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. \end{equation*}
12

From Exercise 17.5.11 obtain the resolvent cubic equation

\begin{equation*} z^3 - pz^2 - 4rz + (4pr - q^2) = 0. \end{equation*}

Solving the resolvent cubic equation, put the equation found in Exercise 17.5.10 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.

  1. \(x^4 - x^2 - 3x + 2 = 0\)

  2. \(x^4 + x^3 - 7 x^2 - x + 6 = 0\)

  3. \(x^4 -2 x^2 + 4 x -3 = 0\)

  4. \(x^4 - 4 x^3 + 3x^2 - 5x +2 = 0\)