Exercise 1.3.13

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\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Suppose that a quartic polynomial $f = x^4+bx^3+cx^2+dx+e$ in $\R[x]$ has distinct roots $x_1,x_2,x_3,x_4 \in \C$. The discriminant of $f$ is defined by the equation
$$\Delta = (x_1-x_2)^2(x_1-x_3)^2(x_1-x_4)^2(x_2-x_3)^2(x_2-x_4)^2(x_3-x_4)^2.$$
The theory developed in Chapter 2 will imply that $\Delta \in \R$, and $\Delta 
eq 0$, since the $x_i$ are distinct. Adapt the proof of Theorem 1.3.1 to show that

$\Delta<0 \iff x^4+bx^3+cx^2+dx+e = 0$ has exactly two real roots.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\be} {\begin{enumerate}} ewcommand{\ee} {\end{enumerate}} ewcommand{\deb}{\begin{eqnarray*}} ewcommand{\fin}{\end{eqnarray*}} ewcommand{\ssi} {si et seulement si } ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} Suppose that $f$ has exactly 2 real roots $x_1,x_2$. Then the two others form a complex conjugate pair : $x_3 = a+bi, x_4 = a-bi, \ b eq 0$. Then \begin{align*} \Delta &= (x_1-x_2)^2 (x_1-x_3)^2 (x_1 - x_4)^2 (x_2-x_3)^2 (x_2-x_4)^2 (x_3-x_4)^2\ &=(x_1-x_2)^2(x_1-a-bi)^2(x_1-a+bi)^2(x_2-a-bi)^2(x_2-a+bi)^2(2bi)^2\ &=-4b^2(x_1-x_2)^2\vert x_1-a-bi\vert^4 \vert x_2 -a -bi \vert^4 <0. \end{align*} The only other possible cases are the case where the 4 roots are real, and then $\Delta >0$, and the case where the 4 roots are non real, forming two complex conjugate pairs. $x_1 = a+ib, x_2 = a-ib, x_3= c+id, x_4 = c-id,\ a,b,c,d\in \mathbb{R}$. Then \begin{align*} \Delta &= (x_1-x_2)^2 (x_1-x_3)^2 (x_1 - x_4)^2 (x_2-x_3)^2 (x_2-x_4)^2 (x_3-x_4)^2\ &=(x_1-x_2)^2 (x_3-x_4)^2(x_1-x_3)^2(\overline{x_1-x_3})^2 (x_1-x_4)^2(\overline{x_1-x_4})^2\ &=(2ib)^2(2id)^2 \vert x_1-x_3 \vert^4\vert x_1-x_4 \vert^4\ &=16b^2d^2 \vert x_1-x_3 \vert^4\vert x_1-x_4 \vert^4 >0.\ \end{align*} Conclusion : $\Delta<0$ if and only if $f$ has exactly two real roots. \end{proof}

Exercise 1.3.13

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Exercise 1.3.13

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