17.6: Additional Exercises- Solving the Cubic and Quartic Equations
Solve the general quadratic equation
\[ ax^2 + bx + c = 0 \nonumber \]
to obtain
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\text{.} \nonumber \]
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.
Show that any cubic equation of the form
\[ x^3 + bx^2 + cx + d = 0 \nonumber \]
can be reduced to the form \(y^3 + py + q = 0\) by making the substitution \(x = y - b/3\text{.}\)
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*}
Make the substitution
\[ y = z - \frac{p}{3 z} \nonumber \]
for \(y\) in the equation \(y^3 + py + q = 0\) and obtain two solutions \(A\) and \(B\) for \(z^3\text{.}\)
Show that the product of the solutions obtained in (4) is \(-p^3/27\text{,}\) deducing that \(\sqrt[3]{A B} = -p/3\text{.}\)
Prove that the possible solutions for \(z\) in (4) are given by
\[ \sqrt[3]{A}, \quad \omega \sqrt[3]{A}, \quad \omega^2 \sqrt[3]{A}, \quad \sqrt[3]{B}, \quad \omega \sqrt[3]{B}, \quad \omega^2 \sqrt[3]{B} \nonumber \]
and use this result to show that the three possible solutions for \(y\) are
\[ \omega^i \sqrt[3]{-\frac{q}{2}+ \sqrt{\ \frac{p^3}{27} + \frac{q^2}{4}} } + \omega^{2i} \sqrt[3]{-\frac{q}{2}- \sqrt{\ \frac{p^3}{27} + \frac{q^2}{4}} }\text{,} \nonumber \]
where \(i = 0, 1, 2\text{.}\)
The discriminant of the cubic equation is
\[ \Delta = \frac{p^3}{27} + \frac{q^2}{4}\text{.} \nonumber \]
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 \(\Delta \lt 0\text{.}\)
Solve the following cubic equations.
- \(\displaystyle x^3 - 4x^2 + 11 x + 30 = 0\)
- \(\displaystyle x^3 - 3x +5 = 0\)
- \(\displaystyle x^3 - 3x +2 = 0\)
- \(\displaystyle x^3 + x + 3 = 0\)
Show that the general quartic equation
\[ x^4 + ax^3 + bx^2 + cx + d = 0 \nonumber \]
can be reduced to
\[ y^4 + py^2 + qy + r = 0 \nonumber \]
by using the substitution \(x = y - a/4\text{.}\)
Show that
\[ \left( y^2 + \frac{1}{2} z \right)^2 = (z - p)y^2 - qy + \left( \frac{1}{4} z^2 - r \right)\text{.} \nonumber \]
Show that the right-hand side of Exercise \(17.6.10\) can be put in the form \((my + k)^2\) if and only if
\[ q^2 - 4(z - p)\left( \frac{1}{4} z^2 - r \right) = 0\text{.} \nonumber \]
From Exercise \(17.6.11\) obtain the resolvent cubic equation
\[ z^3 - pz^2 - 4rz + (4pr - q^2) = 0\text{.} \nonumber \]
Solving the resolvent cubic equation, put the equation found in Exercise 17.6.10 in the form
\[ \left( y^2 + \frac{1}{2} z \right)^2 = (my + k)^2 \nonumber \]
to obtain the solution of the quartic equation.
Use this method to solve the following quartic equations.
- \(\displaystyle x^4 - x^2 - 3x + 2 = 0\)
- \(\displaystyle x^4 + x^3 - 7 x^2 - x + 6 = 0\)
- \(\displaystyle x^4 -2 x^2 + 4 x -3 = 0\)
- \(\displaystyle x^4 - 4 x^3 + 3x^2 - 5x +2 = 0\)