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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle - 2 x^{2} - 18 x + 20=0\)
Since the GCF is \(\displaystyle -2\) we need we factor out the GCF to get \(\displaystyle -2(x^{2} + 9 x - 10)\). This is now a pq method factoring. The factors of \(\displaystyle -10\) that add up to \(\displaystyle 9\) are \(\displaystyle -1\) and \(\displaystyle 10\). This gives \(\displaystyle -2(x - 1)(x + 10)=0\). The solutions are \(\displaystyle x = 1\) and \(\displaystyle x = -10\)
\begin{question}Solve $- 2 x^{2} - 18 x + 20=0$
\soln{9cm}{Since the GCF is $-2$ we need we factor out the GCF to get $-2(x^{2} + 9 x - 10)$. This is now a pq method factoring. The factors of $-10$ that add up to $9$ are $-1$ and $10$. This gives $-2(x - 1)(x + 10)=0$. The solutions are $x = 1$ and $x = -10$}
\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 - 2 x^{2} - 18 x + 20=0 " src="/equation_images/%20%5Cdisplaystyle%20-%202%20x%5E%7B2%7D%20-%2018%20x%20%2B%2020%3D0%20" alt="LaTeX: \displaystyle - 2 x^{2} - 18 x + 20=0 " data-equation-content=" \displaystyle - 2 x^{2} - 18 x + 20=0 " /> </p> </p><p> <p>Since the GCF is <img class="equation_image" title=" \displaystyle -2 " src="/equation_images/%20%5Cdisplaystyle%20-2%20" alt="LaTeX: \displaystyle -2 " data-equation-content=" \displaystyle -2 " /> we need we factor out the GCF to get <img class="equation_image" title=" \displaystyle -2(x^{2} + 9 x - 10) " src="/equation_images/%20%5Cdisplaystyle%20-2%28x%5E%7B2%7D%20%2B%209%20x%20-%2010%29%20" alt="LaTeX: \displaystyle -2(x^{2} + 9 x - 10) " data-equation-content=" \displaystyle -2(x^{2} + 9 x - 10) " /> . This is now a pq method factoring. The factors of <img class="equation_image" title=" \displaystyle -10 " src="/equation_images/%20%5Cdisplaystyle%20-10%20" alt="LaTeX: \displaystyle -10 " data-equation-content=" \displaystyle -10 " /> that add up to <img class="equation_image" title=" \displaystyle 9 " src="/equation_images/%20%5Cdisplaystyle%209%20" alt="LaTeX: \displaystyle 9 " data-equation-content=" \displaystyle 9 " /> are <img class="equation_image" title=" \displaystyle -1 " src="/equation_images/%20%5Cdisplaystyle%20-1%20" alt="LaTeX: \displaystyle -1 " data-equation-content=" \displaystyle -1 " /> and <img class="equation_image" title=" \displaystyle 10 " src="/equation_images/%20%5Cdisplaystyle%2010%20" alt="LaTeX: \displaystyle 10 " data-equation-content=" \displaystyle 10 " /> . This gives <img class="equation_image" title=" \displaystyle -2(x - 1)(x + 10)=0 " src="/equation_images/%20%5Cdisplaystyle%20-2%28x%20-%201%29%28x%20%2B%2010%29%3D0%20" alt="LaTeX: \displaystyle -2(x - 1)(x + 10)=0 " data-equation-content=" \displaystyle -2(x - 1)(x + 10)=0 " /> . The solutions are <img class="equation_image" title=" \displaystyle x = 1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%201%20" alt="LaTeX: \displaystyle x = 1 " data-equation-content=" \displaystyle x = 1 " /> and <img class="equation_image" title=" \displaystyle x = -10 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-10%20" alt="LaTeX: \displaystyle x = -10 " data-equation-content=" \displaystyle x = -10 " /> </p> </p>