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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle 5 x^{2} + 5 x - 100=0\)
Since the GCF is \(\displaystyle 5\) we need we factor out the GCF to get \(\displaystyle 5(x^{2} + x - 20)\). This is now a pq method factoring. The factors of \(\displaystyle -20\) that add up to \(\displaystyle 1\) are \(\displaystyle 5\) and \(\displaystyle -4\). This gives \(\displaystyle 5(x + 5)(x - 4)=0\). The solutions are \(\displaystyle x = -5\) and \(\displaystyle x = 4\)
\begin{question}Solve $5 x^{2} + 5 x - 100=0$
\soln{9cm}{Since the GCF is $5$ we need we factor out the GCF to get $5(x^{2} + x - 20)$. This is now a pq method factoring. The factors of $-20$ that add up to $1$ are $5$ and $-4$. This gives $5(x + 5)(x - 4)=0$. The solutions are $x = -5$ and $x = 4$}
\end{question}
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\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 5 x^{2} + 5 x - 100=0 " src="/equation_images/%20%5Cdisplaystyle%205%20x%5E%7B2%7D%20%2B%205%20x%20-%20100%3D0%20" alt="LaTeX: \displaystyle 5 x^{2} + 5 x - 100=0 " data-equation-content=" \displaystyle 5 x^{2} + 5 x - 100=0 " /> </p> </p><p> <p>Since the GCF is <img class="equation_image" title=" \displaystyle 5 " src="/equation_images/%20%5Cdisplaystyle%205%20" alt="LaTeX: \displaystyle 5 " data-equation-content=" \displaystyle 5 " /> we need we factor out the GCF to get <img class="equation_image" title=" \displaystyle 5(x^{2} + x - 20) " src="/equation_images/%20%5Cdisplaystyle%205%28x%5E%7B2%7D%20%2B%20x%20-%2020%29%20" alt="LaTeX: \displaystyle 5(x^{2} + x - 20) " data-equation-content=" \displaystyle 5(x^{2} + x - 20) " /> . This is now a pq method factoring. The factors of <img class="equation_image" title=" \displaystyle -20 " src="/equation_images/%20%5Cdisplaystyle%20-20%20" alt="LaTeX: \displaystyle -20 " data-equation-content=" \displaystyle -20 " /> that add up to <img class="equation_image" title=" \displaystyle 1 " src="/equation_images/%20%5Cdisplaystyle%201%20" alt="LaTeX: \displaystyle 1 " data-equation-content=" \displaystyle 1 " /> are <img class="equation_image" title=" \displaystyle 5 " src="/equation_images/%20%5Cdisplaystyle%205%20" alt="LaTeX: \displaystyle 5 " data-equation-content=" \displaystyle 5 " /> and <img class="equation_image" title=" \displaystyle -4 " src="/equation_images/%20%5Cdisplaystyle%20-4%20" alt="LaTeX: \displaystyle -4 " data-equation-content=" \displaystyle -4 " /> . This gives <img class="equation_image" title=" \displaystyle 5(x + 5)(x - 4)=0 " src="/equation_images/%20%5Cdisplaystyle%205%28x%20%2B%205%29%28x%20-%204%29%3D0%20" alt="LaTeX: \displaystyle 5(x + 5)(x - 4)=0 " data-equation-content=" \displaystyle 5(x + 5)(x - 4)=0 " /> . The solutions are <img class="equation_image" title=" \displaystyle x = -5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-5%20" alt="LaTeX: \displaystyle x = -5 " data-equation-content=" \displaystyle x = -5 " /> and <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX: \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " /> </p> </p>