17.6: Additional Exercises- Solving the Cubic and Quartic Equations
- Page ID
- 81176
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
- \(\displaystyle x^4 - x^2 - 3x + 2 = 0\)
- \(\displaystyle x^4 + x^3 - 7 x^2 - x + 6 = 0\)
- \(\displaystyle x^4 -2 x^2 + 4 x -3 = 0\)
- \(\displaystyle x^4 - 4 x^3 + 3x^2 - 5x +2 = 0\)