Exercises6.8Additional Exercises: Solving the Cubic and Quartic Equations

1.

\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

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 $y^3 + py + q=0$

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. $\displaystyle x^3 - 4x^2 + 11 x + 30 = 0$

2. $\displaystyle x^3 - 3x +5 = 0$

3. $\displaystyle x^3 - 3x +2 = 0$

4. $\displaystyle x^3 + x + 3 = 0$

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 6.8.10 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.

From Exercise 6.8.11 obtain the resolvent cubic equation

\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 6.8.10 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. $\displaystyle x^4 - x^2 - 3x + 2 = 0$

2. $\displaystyle x^4 + x^3 - 7 x^2 - x + 6 = 0$

3. $\displaystyle x^4 -2 x^2 + 4 x -3 = 0$

4. $\displaystyle x^4 - 4 x^3 + 3x^2 - 5x +2 = 0$