Exercises 17.6 Additional Exercises: Solving the Cubic and Quartic Equations
1.
Complete the square to solve the general quadratic equation
to obtain
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
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
4.
Make the substitution
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 \(\sqrt[3]{A B} = -p/3\text{.}\)
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
where \(i = 0, 1, 2\text{.}\)
7.
The discriminant of the cubic equation is
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 \(\Delta \lt 0\text{.}\)
8.
Solve the following cubic equations.
\(\displaystyle x^3 - 4x^2 + 11 x + 30 = 0\)
\(\displaystyle x^3 - 3x +5 = 0\)
\(\displaystyle x^3 - 3x +2 = 0\)
\(\displaystyle x^3 + x + 3 = 0\)
9.
Show that the general quartic equation
can be reduced to
by using the substitution \(x = y - a/4\text{.}\)
10.
Show that
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
12.
From Exercise 17.6.11 obtain the resolvent cubic equation
Solving the resolvent cubic equation, put the equation found in Exercise 17.6.10 in the form
to obtain the solution of the quartic equation.
13.
Use this method to solve the following quartic equations.
\(\displaystyle x^4 - x^2 - 3x + 2 = 0\)
\(\displaystyle x^4 + x^3 - 7 x^2 - x + 6 = 0\)
\(\displaystyle x^4 -2 x^2 + 4 x -3 = 0\)
\(\displaystyle x^4 - 4 x^3 + 3x^2 - 5x +2 = 0\)