{ "cells": [ {"cell_type": "code", "execution_count":null, "metadata":{}, "source":["%%html\n\n\n\n\n\n\n"], "outputs":[]}, {"cell_type":"markdown", "metadata":{}, "source":["**Important:** to view this notebook properly you will need to execute the cell above, which assumes you have an Internet connection. It should already be selected, or place your cursor anywhere above to select. Then press the \"Run\" button in the menu bar above (the right-pointing arrowhead), or press Shift-Enter on your keyboard."]}, {"cell_type":"markdown", "metadata":{}, "source":["$\\newcommand{\\identity}{\\mathrm{id}}\n\\newcommand{\\notdivide}{\\nmid}\n\\newcommand{\\notsubset}{\\not\\subset}\n\\newcommand{\\lcm}{\\operatorname{lcm}}\n\\newcommand{\\gf}{\\operatorname{GF}}\n\\newcommand{\\inn}{\\operatorname{Inn}}\n\\newcommand{\\aut}{\\operatorname{Aut}}\n\\newcommand{\\Hom}{\\operatorname{Hom}}\n\\newcommand{\\cis}{\\operatorname{cis}}\n\\newcommand{\\chr}{\\operatorname{char}}\n\\newcommand{\\Null}{\\operatorname{Null}}\n\\newcommand{\\lt}{<}\n\\newcommand{\\gt}{>}\n\\newcommand{\\amp}{&}\n$"]}, {"cell_type":"markdown", "metadata":{}, "source":["

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

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
1

Solve the general quadratic equation

\n\\begin{equation*}\nax^2 + bx + c = 0\n\\end{equation*}\n

to obtain

\n\\begin{equation*}\nx = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}.\n\\end{equation*}\n

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.

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
2

Show that any cubic equation of the form

\n\\begin{equation*}\nx^3 + bx^2 + cx + d = 0\n\\end{equation*}\n

can be reduced to the form $y^3 + py + q = 0$ by making the substitution $x = y - b/3\\text{.}$

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
3

Prove that the cube roots of 1 are given by

\n\\begin{align*}\n\\omega & = \\frac{-1+ i \\sqrt{3}}{2}\\\\\n\\omega^2 & = \\frac{-1- i \\sqrt{3}}{2}\\\\\n\\omega^3 & = 1.\n\\end{align*}\n
"]}, {"cell_type":"markdown", "metadata":{}, "source":["
4

Make the substitution

\n\\begin{equation*}\ny = z - \\frac{p}{3 z}\n\\end{equation*}\n

for $y$ in the equation $y^3 + py + q = 0$ and obtain two solutions $A$ and $B$ for $z^3\\text{.}$

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
5

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

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
6

Prove that the possible solutions for $z$ in (4) are given by

\n\\begin{equation*}\n\\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}\n\\end{equation*}\n

and use this result to show that the three possible solutions for $y$ are

\n\\begin{equation*}\n\\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}} },\n\\end{equation*}\n

where $i = 0, 1, 2\\text{.}$

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
7

The discriminant of the cubic equation is

\n\\begin{equation*}\n\\Delta = \\frac{p^3}{27} + \\frac{q^2}{4}.\n\\end{equation*}\n

Show that $y^3 + py + q=0$

  1. has three real roots, at least two of which are equal, if $\\Delta = 0\\text{.}$

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

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

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
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. $x^3 + x + 3 = 0$

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
9

Show that the general quartic equation

\n\\begin{equation*}\nx^4 + ax^3 + bx^2 + cx + d = 0\n\\end{equation*}\n

can be reduced to

\n\\begin{equation*}\ny^4 + py^2 + qy + r = 0\n\\end{equation*}\n

by using the substitution $x = y - a/4\\text{.}$

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
10

Show that

\n\\begin{equation*}\n\\left( y^2 + \\frac{1}{2} z \\right)^2 = (z - p)y^2 - qy + \\left( \\frac{1}{4} z^2 - r \\right).\n\\end{equation*}\n
"]}, {"cell_type":"markdown", "metadata":{}, "source":["
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

\n\\begin{equation*}\nq^2 - 4(z - p)\\left( \\frac{1}{4} z^2 - r \\right) = 0.\n\\end{equation*}\n
"]}, {"cell_type":"markdown", "metadata":{}, "source":["
12

From Exercise 17.5.11 obtain the resolvent cubic equation

\n\\begin{equation*}\nz^3 - pz^2 - 4rz + (4pr - q^2) = 0.\n\\end{equation*}\n

Solving the resolvent cubic equation, put the equation found in Exercise 17.5.10 in the form

\n\\begin{equation*}\n\\left( y^2 + \\frac{1}{2} z \\right)^2 = (my + k)^2\n\\end{equation*}\n

to obtain the solution of the quartic equation.

"]}, {"cell_type":"markdown", "metadata":{}, "source":["
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. $x^4 - 4 x^3 + 3x^2 - 5x +2 = 0$

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