在\amsmath
和中\tasks
,提供了求三次方程式 3 根的逐步求解過程。
請您協助解決 2 個問題:
- 將每個問題中 2 個立方根之間的水平間距減小到大約 2-3 毫米:
- 插入標註(左列和右列)以識別產生根的第一個因子。也,將黃色突出顯示的因素加粗。
謝謝你! mwe在下面
\documentclass[12pt]{exam}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{framed} %box para
\usepackage{multicol}
\usepackage{tasks}
\usepackage{xcolor}
\usepackage[margin=0.5in]{geometry}
%\usepackage{bm}%bold equation
\setlength{\parindent}{0pt} % removes paragraph indentation
\pagestyle{head}
\header{Algegra II: Assignment: 10-C Solving Polynomial Equations}
{}
{Due 03/12/2023}
\newcommand{\pagetop}{%
%\makebox[\textwidth]%{Name:\enspace\hrulefill}\par
\vspace{4mm}
\fbox{\fbox{\parbox{\dimexpr\textwidth-4\fboxsep-4\fboxrule}{
\textbf {Solve each polynomial equation by factoring. Find all real and/or imaginary/complex roots. Simplify answers.}
%\par
%\bigskip
}}}\par
\vspace{0.5mm}
}
\setlength{\jot}{1em}
%define highlighting
\newcommand{\hll}[1]{\colorbox{yellow}{$\displaystyle #1$}}
\begin{document}
\pagetop
\settasks{
after-item-skip=3em, after-skip=2cm,
label-width=2em,
item-indent=3em,
label=(\arabic*),
column-sep=2em
}
\begin{tasks}(2)
%Prob #1
\task \(\begin{aligned}[t]
&x^3-216=0 \\
& \hspace{2em}\begin{aligned}[t]
&\sqrt[3]{x^3} =3
&\sqrt[3]{216} =6\\
&\hll {(x-6)}(x^2+6x+6^2)\\
&\hll{(x-6)}(x^2+6x+36)\\
&a = 1; b = 6; c = 36\\
&x=\frac{-(b)\pm\sqrt{(b^2)-4(a)(c)}}{2(a)}\\
&x=\frac{-(6)\pm\sqrt{(6^2)-4(1)(36)}}{2(1)}\\
&x=\frac{-6\pm\sqrt{-144}}{2}\\
&x=\frac{-(6)\pm\sqrt{(6^2)-4(1)(36)}}{2(1)}\\
&x=\frac{-6\pm\sqrt{-108}}{2}\\
&x=\frac{-6\pm 6i\sqrt{3}}{2}\\
&x=\frac{2(-3\pm 3i\sqrt{3})}{2}\\
&x=\-3\pm 3i\sqrt{3}&x=6\\
\end{aligned}
\end{aligned}\)
%Problem #2
\task \(\begin{aligned}[t]
&8x^3 +125 \\
& \hspace{2em}\begin{aligned}[t]
&\sqrt[3]{8x^3} =2x
&\sqrt[3]{125} =5\\
&\hll{(2x+5)}(4x^2-10x+5^2)\\
&\hll{(2x+5)}(4x^2-10x+25)\\
&a = 4; b = -10; c = 25\\
&x=\frac{-(b)\pm\sqrt{(b^2)-4(a)(c)}}{2(a)}\\
&x=\frac{-(-10)\pm\sqrt{(-10^2)-4(4)(25)}}{2(4)}\\
&x=\frac{10\pm\sqrt{-300}}{8}\\
&x=\frac{10\pm 10i\sqrt{3}}{8}\\
&x=\frac{2(5\pm 5i\sqrt{3})}{8}\\
&x=\frac{5\pm 5i\sqrt{3})}{4}&x=-\frac{-5}{2}\\
\end{aligned}
\end{aligned}\)
%Problem #3
\end{tasks}
\end{document}```
答案1
讓我們分部分來說:
對於簡單的間距,我們可以使用命令
\quad
或\qquad
在間距中保持一定的標準化,在您的情況下,使用\qquad
是最好的做法。為了創建帶有警告框的黃色框,我創建了一個新命令來分配兩個條目:
\callout{#1}{#2}
.第一個條目#1
是以黃色突出顯示的方程項,而第二個條目#2
是方程式的其餘部分,不會突出顯示。需要評論的一件重要事情是需要指令,在本例中是[2.2em]
在換行符之前。透過舉例就更清楚了。
套餐
\documentclass[12pt]{exam}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{framed} % box para
\usepackage{multicol}
\usepackage{tasks}
\usepackage{xcolor}
\usepackage[margin=0.5in]{geometry}
\usepackage{tikz}
\usetikzlibrary{shapes.callouts} % To create the call out boxes
新指令
\newcommand*{\callout}[2]{\hspace{-5em}
\tikz[baseline=(X)] \node[shape = rectangle callout,
fill=white,
draw= black,
minimum width=4.5em,
rounded corners,
callout relative pointer={(+0.55,-0.45)},
font = {\sffamily},
] (X) {${\scriptsize\begin{matrix}\textrm{1st root/} \\ \textrm{solution}\end{matrix}}$};%
\tikz[baseline=(X.base)] \node[rectangle, fill=yellow, inner sep=1mm] (Y) at ([yshift = -2.2em]X) {$#1$};%
\tikz[baseline=(X.base)] \node[] at (Y.east) {$#2$};%
}
新指令無垂直間距指令
\begin{tasks}(2)
% Problem 1:
\task \(
\begin{aligned}[t]
& x^3-216 = 0 \\
& \hspace{2em}
\begin{aligned}[t]
& \sqrt[3]{x^3} = 3 \qquad \sqrt[3]{216} = 6 \\
& \hll{(x-6)}(x^2+6x+6^2) \\%[-2.2em]
% Do not space here: ^
& \callout{(x-6)}{(x^2+6x+36)} \\
\end{aligned}
\end{aligned}
\)
\end{tasks}
垂直間距指令內的新指令
但是,如果我們包含這些指令:
\begin{tasks}(2)
% Problem 1:
\task \(
\begin{aligned}[t]
& x^3-216 = 0 \\
& \hspace{2em}
\begin{aligned}[t]
& \sqrt[3]{x^3} = 3 \qquad \sqrt[3]{216} = 6 \\
& \hll{(x-6)}(x^2+6x+6^2) \\[-2.2em]
% Do not space here: ^
& \callout{(x-6)}{(x^2+6x+36)} \\
\end{aligned}
\end{aligned}
\)
\end{tasks}
關於這個新命令的一點評論
我添加了一條註釋,不要將雙欄\\
和指令隔開[-2.2em]
,這是因為這些指令不適用於此空間。
完整程式碼
最後,我將發布所有程式碼和相應的輸出:
\documentclass[12pt]{exam}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{framed} %box para
\usepackage{multicol}
\usepackage{tasks}
\usepackage{xcolor}
\usepackage[margin=0.5in]{geometry}
\usepackage{tikz}
\usetikzlibrary{shapes.callouts}
\newcommand*{\callout}[2]{\hspace{-5em}
\tikz[baseline=(X)] \node[shape = rectangle callout,
fill=white,
draw= black,
minimum width=4.5em,
rounded corners,
callout relative pointer={(+0.55,-0.45)},
font = {\sffamily},
] (X) {${\scriptsize\begin{matrix}\textrm{1st root/} \\ \textrm{solution}\end{matrix}}$};%
\tikz[baseline=(X.base)] \node[rectangle, fill=yellow, inner sep=1mm] (Y) at ([yshift = -2.2em]X) {$#1$};%
\tikz[baseline=(X.base)] \node[] at (Y.east) {$#2$};%
}
%\usepackage{bm}%bold equation
\setlength{\parindent}{0pt} % removes paragraph indentation
\pagestyle{head}
\header{Algebra II: Assignment: 10-C Solving Polynomial Equations}
{}
{Due 03/12/2023}
\newcommand{\pagetop}{%
%\makebox[\textwidth]%{Name:\enspace\hrulefill}\par
\vspace{4mm}
\fbox{\fbox{\parbox{\dimexpr\textwidth-4\fboxsep-4\fboxrule}{
\textbf {Solve each polynomial equation by factoring. Find all real and/or imaginary/complex roots. Simplify answers.}
%\par
%\bigskip
}}}\par
\vspace{0.5mm}
}
\setlength{\jot}{1em}
%define highlighting
\newcommand{\hll}[1]{\colorbox{yellow}{$\displaystyle #1$}}
\begin{document}
\pagetop
\settasks{
after-item-skip=3em, after-skip=2cm,
label-width=2em,
item-indent=3em,
label=(\arabic*),
column-sep=2em
}
\begin{tasks}(2)
% Problem 1:
\task \(
\begin{aligned}[t]
& x^3-216 = 0 \\
& \hspace{2em}
\begin{aligned}[t]
& \sqrt[3]{x^3} = 3 \qquad \sqrt[3]{216} = 6 \\
& \hll{(x-6)}(x^2+6x+6^2) \\[-2.2em]
% Do not space here: ^
& \callout{(x-6)}{(x^2+6x+36)} \\
& a = 1; b = 6; c = 36 \\
& x = \frac{-(b)\pm\sqrt{(b^2)-4(a)(c)}}{2(a)} \\
& x = \frac{-(6)\pm\sqrt{(6^2)-4(1)(36)}}{2(1)} \\
& x = \frac{-6\pm\sqrt{-144}}{2} \\
& x = \frac{-(6)\pm\sqrt{(6^2)-4(1)(36)}}{2(1)} \\
& x = \frac{-6\pm\sqrt{-108}}{2} \\
& x = \frac{-6\pm 6i\sqrt{3}}{2} \\
& x = \frac{2(-3\pm 3i\sqrt{3})}{2} \\
& x = \-3\pm 3i\sqrt{3} \qquad x=6 \\
\end{aligned}
\end{aligned}
\)
% Problem #2
\task
\(\begin{aligned}[t]
& 8x^3 +125 \\
& \hspace{2em}
\begin{aligned}[t]
& \sqrt[3]{8x^3} = 2x \qquad \sqrt[3]{125} = 5 \\
& \hll{(2x+5)}(4x^2-10x+5^2) \\[-2.2em]
& \callout{(2x+5)}{(4x^2-10x+25)} \\
& a = 4; b = -10; c = 25 \\
& x = \frac{-(b)\pm\sqrt{(b^2)-4(a)(c)}}{2(a)} \\
& x = \frac{-(-10)\pm\sqrt{(-10^2)-4(4)(25)}}{2(4)} \\
& x = \frac{10\pm\sqrt{-300}}{8} \\
& x = \frac{10\pm 10i\sqrt{3}}{8} \\
& x = \frac{2(5\pm 5i\sqrt{3})}{8} \\
& x = \frac{5\pm 5i\sqrt{3})}{4} \qquad x=-\frac{-5}{2} \\
\end{aligned}
\end{aligned}
\)
% Problem 3
\end{tasks}
\end{document}