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Section 17.5 Additional Exercises: Solving the Cubic and Quartic Equations

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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.

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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 \(x = y - b/3\text{.}\)

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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*}

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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 \(z^3\text{.}\)

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5

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

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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 \(i = 0, 1, 2\text{.}\)

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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 \(\Delta = 0\text{.}\)

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

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

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8

Solve the following cubic equations.

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

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

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

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

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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 \(x = y - a/4\text{.}\)

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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*}

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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*}

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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.

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13

Use this method to solve the following quartic equations.

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

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

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

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