\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

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2017.3.3 Question 3

By Vieta’s Theorem, from the quartic equation in \(x\), we have \[ \alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = q, \] and from the cubic equation in \(y\), we have \[ \left (\alpha \beta + \gamma \delta \right ) + \left (\alpha \gamma + \beta \delta \right ) + \left (\alpha \delta + \beta \gamma \right ) = -A. \]

Therefore, \(A = -q\).

  1. Since \((p, q, r, s) = (0, 3, -6, 10)\), the cubic equation is reduced to \[ y^3 - 3y^2 - 10y + 84 = 0, \] and therefore \[ (y - 2)(y - 7)(y + 6) = 0. \] Therefore, \(y_1 = 7, y_2 = 2, y_3 = -6\), and \(\alpha \beta + \gamma \delta = 7\).

  2. We have \begin {align*} (\alpha + \beta )(\gamma + \delta ) & = \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta \\ & = \left (\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta \right ) - \left (\alpha \beta + \gamma \delta \right ) \\ & = q - 7 \\ & = 3 - 7 \\ & = -4. \end {align*}

    By Vieta’s Theorem, we have \(\alpha \beta \gamma \delta = s = 10\). Therefore, \(\alpha \beta \) and \(\gamma \delta \) must be roots to the equation \[ x^2 - 7x + 10 = 0. \]

    The two roots are \(x = 2\) and \(x = 5\), and therefore \(\alpha \beta = 5\).

  3. We have from the other root that \(\gamma \delta = 2\).

    We notice that \((\alpha + \beta ) + (\gamma + \delta ) = -p = 0\). Therefore, from part 2, \((\alpha + \beta )\) and \((\gamma + \delta )\) are roots to the equation \[ x^2 - 4 = 0. \]

    This gives us \(\alpha + \beta = \pm 2\) and \(\gamma + \delta = \mp 2\).

    Using the value of \(r\) and Vieta’s Theorem, we have \[ \alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = -r = 6. \]

    Plugging in \(\alpha \beta = 5\) and \(\gamma \delta = 2\), we have \[ 5(\gamma + \delta ) + 2(\alpha + \beta ) = 6. \]

    Therefore, it must be the case that \(\alpha + \beta = -2\) and \(\gamma + \delta = 2\).

    Hence, using the values of \(\alpha \beta \) and \(\gamma \delta \), \(\alpha \) and \(\beta \) are solutions to the quadratic equation \(x^2 + 2x + 5 = 0\), and \(\gamma \) and \(\delta \) are solutions to the quadratic equation \(x^2 - 2x + 2 = 0\).

    Solving this gives us \(\alpha , \beta = -1 \pm 2i\) and \(\gamma , \delta = 1 \pm i\). The solutions to the original quartic equation is \[ x_{1, 2} = -1 \pm 2i, x_{3, 4} = 1 \pm i. \]

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