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

##### 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 \(y^3 + py + q=0\)

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

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

has three distinct real roots if \(\Delta \lt 0\).

##### 8

Solve the following cubic equations.

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

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

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

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

##### 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.

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

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

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

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