##### 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 $\Delta \gt 0$, the equation has two distinct real solutions. If $\Delta = 0$, the equation has a single repeated real root. If $\Delta \lt 0$, 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 $x = y - b/3$.

##### 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 $z^3$.

##### 5

Show that the product of the solutions obtained in (4) is $-p^3/27$, deducing that $\sqrt[3]{A B} = -p/3$.

##### 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 $i = 0, 1, 2$.

##### 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 $\Delta = 0$.

2. has one real root and two conjugate imaginary roots if $\Delta \gt 0$.

3. has three distinct real roots if $\Delta \lt 0$.

##### 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 $x = y - a/4$.

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