\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Solve the inequality \(\displaystyle \frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5}\)
Getting zero on one side and factoring gives \(\displaystyle - \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0\). This gives the least common denominator as \(\displaystyle \left(x - 5\right) \left(x - 1\right) \left(x + 5\right)\). Building each fraction to get the common denominator gives \(\displaystyle \frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0\). Simplifying gives \(\displaystyle \frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0\). The inequality can change signs at the zeros of the numerator, \(\displaystyle \emptyset\), or the zeros of the denominator \(\displaystyle \left\{-5, 1, 5\right\}\). Making a sign chart gives: This gives the solution \(\displaystyle \left(-\infty, -5\right) \cup \left(1, 5\right)\)
\begin{question}Solve the inequality $\frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5}$
\soln{9cm}{Getting zero on one side and factoring gives $- \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0$. This gives the least common denominator as $\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)$. Building each fraction to get the common denominator gives $\frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0$. Simplifying gives $\frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0$. The inequality can change signs at the zeros of the numerator, $\emptyset$, or the zeros of the denominator $\left\{-5, 1, 5\right\}$. Making a sign chart gives: \begin{tikzpicture}
\draw[latex-latex, thick] (-3, 0) -- (5, 0);
\draw(-1, 0.2) -- (-1, -0.2);
\draw (-1,-0.2) node[below]{$-5$};
\draw (-2,.2) node[above]{$-$};
\draw(1, 0.2) -- (1, -0.2);
\draw (1,-0.2) node[below]{$1$};
\draw (0,.2) node[above]{$+$};
\draw(3, 0.2) -- (3, -0.2);
\draw (3,-0.2) node[below]{$5$};
\draw (4,.2) node[above]{$+$};
\draw (2,.2) node[above]{$-$};
\end{tikzpicture}
This gives the solution $\left(-\infty, -5\right) \cup \left(1, 5\right)$}
\end{question}
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}
\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}
\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}
\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Solve the inequality <img class="equation_image" title=" \displaystyle \frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%20-%2025%7D%3C%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%20%2B%204%20x%20-%205%7D%20" alt="LaTeX: \displaystyle \frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5} " data-equation-content=" \displaystyle \frac{2}{x^{2} - 25}<\frac{2}{x^{2} + 4 x - 5} " /> </p> </p><p> <p>Getting zero on one side and factoring gives <img class="equation_image" title=" \displaystyle - \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0 " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B2%7D%7B%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%20%2B%20%5Cfrac%7B2%7D%7B%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%3C%200%20" alt="LaTeX: \displaystyle - \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0 " data-equation-content=" \displaystyle - \frac{2}{\left(x - 1\right) \left(x + 5\right)} + \frac{2}{\left(x - 5\right) \left(x + 5\right)}< 0 " /> . This gives the least common denominator as <img class="equation_image" title=" \displaystyle \left(x - 5\right) \left(x - 1\right) \left(x + 5\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%20" alt="LaTeX: \displaystyle \left(x - 5\right) \left(x - 1\right) \left(x + 5\right) " data-equation-content=" \displaystyle \left(x - 5\right) \left(x - 1\right) \left(x + 5\right) " /> . Building each fraction to get the common denominator gives <img class="equation_image" title=" \displaystyle \frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B2%20x%20-%202%20-%20%282%20x%20-%2010%29%7D%7B%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%20%3C%200%20" alt="LaTeX: \displaystyle \frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0 " data-equation-content=" \displaystyle \frac{2 x - 2 - (2 x - 10)}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)} < 0 " /> . Simplifying gives <img class="equation_image" title=" \displaystyle \frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B8%7D%7B%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28x%20-%201%5Cright%29%20%5Cleft%28x%20%2B%205%5Cright%29%7D%3C0%20" alt="LaTeX: \displaystyle \frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0 " data-equation-content=" \displaystyle \frac{8}{\left(x - 5\right) \left(x - 1\right) \left(x + 5\right)}<0 " /> . The inequality can change signs at the zeros of the numerator, <img class="equation_image" title=" \displaystyle \emptyset " src="/equation_images/%20%5Cdisplaystyle%20%5Cemptyset%20" alt="LaTeX: \displaystyle \emptyset " data-equation-content=" \displaystyle \emptyset " /> , or the zeros of the denominator <img class="equation_image" title=" \displaystyle \left\{-5, 1, 5\right\} " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%5C%7B-5%2C%201%2C%205%5Cright%5C%7D%20" alt="LaTeX: \displaystyle \left\{-5, 1, 5\right\} " data-equation-content=" \displaystyle \left\{-5, 1, 5\right\} " /> . Making a sign chart gives: <?xml version="1.0" encoding="UTF-8"?>
<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" width="227.569pt" height="38.911pt" viewBox="0 0 227.569 38.911" version="1.1">
<defs>
<g>
<symbol overflow="visible" id="glyph0-0">
<path style="stroke:none;" d=""/>
</symbol>
<symbol overflow="visible" id="glyph0-1">
<path style="stroke:none;" d="M 6.5625 -2.296875 C 6.734375 -2.296875 6.921875 -2.296875 6.921875 -2.5 C 6.921875 -2.6875 6.734375 -2.6875 6.5625 -2.6875 L 1.171875 -2.6875 C 1 -2.6875 0.828125 -2.6875 0.828125 -2.5 C 0.828125 -2.296875 1 -2.296875 1.171875 -2.296875 Z M 6.5625 -2.296875 "/>
</symbol>
<symbol overflow="visible" id="glyph1-0">
<path style="stroke:none;" d=""/>
</symbol>
<symbol overflow="visible" id="glyph1-1">
<path style="stroke:none;" d="M 4.46875 -2 C 4.46875 -3.1875 3.65625 -4.1875 2.578125 -4.1875 C 2.109375 -4.1875 1.671875 -4.03125 1.3125 -3.671875 L 1.3125 -5.625 C 1.515625 -5.5625 1.84375 -5.5 2.15625 -5.5 C 3.390625 -5.5 4.09375 -6.40625 4.09375 -6.53125 C 4.09375 -6.59375 4.0625 -6.640625 3.984375 -6.640625 C 3.984375 -6.640625 3.953125 -6.640625 3.90625 -6.609375 C 3.703125 -6.515625 3.21875 -6.3125 2.546875 -6.3125 C 2.15625 -6.3125 1.6875 -6.390625 1.21875 -6.59375 C 1.140625 -6.625 1.125 -6.625 1.109375 -6.625 C 1 -6.625 1 -6.546875 1 -6.390625 L 1 -3.4375 C 1 -3.265625 1 -3.1875 1.140625 -3.1875 C 1.21875 -3.1875 1.234375 -3.203125 1.28125 -3.265625 C 1.390625 -3.421875 1.75 -3.96875 2.5625 -3.96875 C 3.078125 -3.96875 3.328125 -3.515625 3.40625 -3.328125 C 3.5625 -2.953125 3.59375 -2.578125 3.59375 -2.078125 C 3.59375 -1.71875 3.59375 -1.125 3.34375 -0.703125 C 3.109375 -0.3125 2.734375 -0.0625 2.28125 -0.0625 C 1.5625 -0.0625 0.984375 -0.59375 0.8125 -1.171875 C 0.84375 -1.171875 0.875 -1.15625 0.984375 -1.15625 C 1.3125 -1.15625 1.484375 -1.40625 1.484375 -1.640625 C 1.484375 -1.890625 1.3125 -2.140625 0.984375 -2.140625 C 0.84375 -2.140625 0.5 -2.0625 0.5 -1.609375 C 0.5 -0.75 1.1875 0.21875 2.296875 0.21875 C 3.453125 0.21875 4.46875 -0.734375 4.46875 -2 Z M 4.46875 -2 "/>
</symbol>
<symbol overflow="visible" id="glyph1-2">
<path style="stroke:none;" d="M 2.9375 -6.375 C 2.9375 -6.625 2.9375 -6.640625 2.703125 -6.640625 C 2.078125 -6 1.203125 -6 0.890625 -6 L 0.890625 -5.6875 C 1.09375 -5.6875 1.671875 -5.6875 2.1875 -5.953125 L 2.1875 -0.78125 C 2.1875 -0.421875 2.15625 -0.3125 1.265625 -0.3125 L 0.953125 -0.3125 L 0.953125 0 C 1.296875 -0.03125 2.15625 -0.03125 2.5625 -0.03125 C 2.953125 -0.03125 3.828125 -0.03125 4.171875 0 L 4.171875 -0.3125 L 3.859375 -0.3125 C 2.953125 -0.3125 2.9375 -0.421875 2.9375 -0.78125 Z M 2.9375 -6.375 "/>
</symbol>
<symbol overflow="visible" id="glyph1-3">
<path style="stroke:none;" d="M 4.078125 -2.296875 L 6.859375 -2.296875 C 7 -2.296875 7.1875 -2.296875 7.1875 -2.5 C 7.1875 -2.6875 7 -2.6875 6.859375 -2.6875 L 4.078125 -2.6875 L 4.078125 -5.484375 C 4.078125 -5.625 4.078125 -5.8125 3.875 -5.8125 C 3.671875 -5.8125 3.671875 -5.625 3.671875 -5.484375 L 3.671875 -2.6875 L 0.890625 -2.6875 C 0.75 -2.6875 0.5625 -2.6875 0.5625 -2.5 C 0.5625 -2.296875 0.75 -2.296875 0.890625 -2.296875 L 3.671875 -2.296875 L 3.671875 0.5 C 3.671875 0.640625 3.671875 0.828125 3.875 0.828125 C 4.078125 0.828125 4.078125 0.640625 4.078125 0.5 Z M 4.078125 -2.296875 "/>
</symbol>
</g>
<clipPath id="clip1">
<path d="M 221 17 L 227.570312 17 L 227.570312 22 L 221 22 Z M 221 17 "/>
</clipPath>
</defs>
<g id="surface1">
<path style="fill:none;stroke-width:0.79701;stroke-linecap:butt;stroke-linejoin:miter;stroke:rgb(0%,0%,0%);stroke-opacity:1;stroke-miterlimit:10;" d="M -80.379406 -0.00134375 L 137.069813 -0.00134375 " transform="matrix(1,0,0,-1,85.438,19.151)"/>
<path style=" stroke:none;fill-rule:nonzero;fill:rgb(0%,0%,0%);fill-opacity:1;" d="M 0.398438 19.152344 C 1.777344 19.410156 4.023438 20.1875 5.578125 21.09375 L 5.578125 17.207031 C 4.023438 18.113281 1.777344 18.890625 0.398438 19.152344 "/>
<g clip-path="url(#clip1)" clip-rule="nonzero">
<path style=" stroke:none;fill-rule:nonzero;fill:rgb(0%,0%,0%);fill-opacity:1;" d="M 227.171875 19.152344 C 225.789062 18.890625 223.546875 18.113281 221.992188 17.207031 L 221.992188 21.09375 C 223.546875 20.1875 225.789062 19.410156 227.171875 19.152344 "/>
</g>
<path style="fill:none;stroke-width:0.3985;stroke-linecap:butt;stroke-linejoin:miter;stroke:rgb(0%,0%,0%);stroke-opacity:1;stroke-miterlimit:10;" d="M -28.348156 5.670531 L -28.348156 -5.669313 " transform="matrix(1,0,0,-1,85.438,19.151)"/>
<g style="fill:rgb(0%,0%,0%);fill-opacity:1;">
<use xlink:href="#glyph0-1" x="50.726" y="34.76"/>
</g>
<g style="fill:rgb(0%,0%,0%);fill-opacity:1;">
<use xlink:href="#glyph1-1" x="58.475" y="34.76"/>
</g>
<g style="fill:rgb(0%,0%,0%);fill-opacity:1;">
<use xlink:href="#glyph0-1" x="24.871" y="9.132"/>
</g>
<path style="fill:none;stroke-width:0.3985;stroke-linecap:butt;stroke-linejoin:miter;stroke:rgb(0%,0%,0%);stroke-opacity:1;stroke-miterlimit:10;" d="M 28.347156 5.670531 L 28.347156 -5.669313 " transform="matrix(1,0,0,-1,85.438,19.151)"/>
<g style="fill:rgb(0%,0%,0%);fill-opacity:1;">
<use xlink:href="#glyph1-2" x="111.294" y="34.76"/>
</g>
<g style="fill:rgb(0%,0%,0%);fill-opacity:1;">
<use xlink:href="#glyph1-3" x="81.563" y="9.132"/>
</g>
<path style="fill:none;stroke-width:0.3985;stroke-linecap:butt;stroke-linejoin:miter;stroke:rgb(0%,0%,0%);stroke-opacity:1;stroke-miterlimit:10;" d="M 85.038562 5.670531 L 85.038562 -5.669313 " transform="matrix(1,0,0,-1,85.438,19.151)"/>
<g style="fill:rgb(0%,0%,0%);fill-opacity:1;">
<use xlink:href="#glyph1-1" x="167.987" y="34.76"/>
</g>
<g style="fill:rgb(0%,0%,0%);fill-opacity:1;">
<use xlink:href="#glyph1-3" x="194.949" y="9.132"/>
</g>
<g style="fill:rgb(0%,0%,0%);fill-opacity:1;">
<use xlink:href="#glyph0-1" x="138.256" y="9.132"/>
</g>
</g>
</svg>
This gives the solution <img class="equation_image" title=" \displaystyle \left(-\infty, -5\right) \cup \left(1, 5\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28-%5Cinfty%2C%20-5%5Cright%29%20%5Ccup%20%5Cleft%281%2C%205%5Cright%29%20" alt="LaTeX: \displaystyle \left(-\infty, -5\right) \cup \left(1, 5\right) " data-equation-content=" \displaystyle \left(-\infty, -5\right) \cup \left(1, 5\right) " /> </p> </p>