\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Solve \(\displaystyle 10 x^{2} + 31 x + 24=0\)
This is an ac method factoring. The factors of \(\displaystyle ac = 240\) that add up to \(\displaystyle 31\) are \(\displaystyle 15\) and \(\displaystyle 16\). Separating gives \(\displaystyle (10 x^{2} + 15 x)+(16 x + 24)=0\). This gives \(\displaystyle (5 x + 8)(2 x + 3)=0\). The solutions are \(\displaystyle x = - \frac{8}{5}\) and \(\displaystyle x = - \frac{3}{2}\)
\begin{question}Solve $10 x^{2} + 31 x + 24=0$
\soln{9cm}{This is an ac method factoring. The factors of $ac = 240$ that add up to $31$ are $15$ and $16$. Separating gives $(10 x^{2} + 15 x)+(16 x + 24)=0$. This gives $(5 x + 8)(2 x + 3)=0$. The solutions are $x = - \frac{8}{5}$ and $x = - \frac{3}{2}$}
\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 <img class="equation_image" title=" \displaystyle 10 x^{2} + 31 x + 24=0 " src="/equation_images/%20%5Cdisplaystyle%2010%20x%5E%7B2%7D%20%2B%2031%20x%20%2B%2024%3D0%20" alt="LaTeX: \displaystyle 10 x^{2} + 31 x + 24=0 " data-equation-content=" \displaystyle 10 x^{2} + 31 x + 24=0 " /> </p> </p><p> <p>This is an ac method factoring. The factors of <img class="equation_image" title=" \displaystyle ac = 240 " src="/equation_images/%20%5Cdisplaystyle%20ac%20%3D%20240%20" alt="LaTeX: \displaystyle ac = 240 " data-equation-content=" \displaystyle ac = 240 " /> that add up to <img class="equation_image" title=" \displaystyle 31 " src="/equation_images/%20%5Cdisplaystyle%2031%20" alt="LaTeX: \displaystyle 31 " data-equation-content=" \displaystyle 31 " /> are <img class="equation_image" title=" \displaystyle 15 " src="/equation_images/%20%5Cdisplaystyle%2015%20" alt="LaTeX: \displaystyle 15 " data-equation-content=" \displaystyle 15 " /> and <img class="equation_image" title=" \displaystyle 16 " src="/equation_images/%20%5Cdisplaystyle%2016%20" alt="LaTeX: \displaystyle 16 " data-equation-content=" \displaystyle 16 " /> . Separating gives <img class="equation_image" title=" \displaystyle (10 x^{2} + 15 x)+(16 x + 24)=0 " src="/equation_images/%20%5Cdisplaystyle%20%2810%20x%5E%7B2%7D%20%2B%2015%20x%29%2B%2816%20x%20%2B%2024%29%3D0%20" alt="LaTeX: \displaystyle (10 x^{2} + 15 x)+(16 x + 24)=0 " data-equation-content=" \displaystyle (10 x^{2} + 15 x)+(16 x + 24)=0 " /> . This gives <img class="equation_image" title=" \displaystyle (5 x + 8)(2 x + 3)=0 " src="/equation_images/%20%5Cdisplaystyle%20%285%20x%20%2B%208%29%282%20x%20%2B%203%29%3D0%20" alt="LaTeX: \displaystyle (5 x + 8)(2 x + 3)=0 " data-equation-content=" \displaystyle (5 x + 8)(2 x + 3)=0 " /> . The solutions are <img class="equation_image" title=" \displaystyle x = - \frac{8}{5} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-%20%5Cfrac%7B8%7D%7B5%7D%20" alt="LaTeX: \displaystyle x = - \frac{8}{5} " data-equation-content=" \displaystyle x = - \frac{8}{5} " /> and <img class="equation_image" title=" \displaystyle x = - \frac{3}{2} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-%20%5Cfrac%7B3%7D%7B2%7D%20" alt="LaTeX: \displaystyle x = - \frac{3}{2} " data-equation-content=" \displaystyle x = - \frac{3}{2} " /> </p> </p>