## Exercises17.6Additional 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 $$\sqrt{A B} = -p/3\text{.}$$

### 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{-\frac{q}{2}+ \sqrt{\ \frac{p^3}{27} + \frac{q^2}{4}} } + \omega^{2i} \sqrt{-\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 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.

From Exercise 17.6.11 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 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.

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