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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle - 20 x^{2} - 39 x - 18=0\)
Since the lead coefficient is negative factor out \(\displaystyle -1\). This is an ac method factoring. The factors of \(\displaystyle ac = 360\) that add up to \(\displaystyle 39\) are \(\displaystyle 24\) and \(\displaystyle 15\). Separating gives \(\displaystyle (20 x^{2} + 24 x)+(15 x + 18)=0\). This gives \(\displaystyle -1(4 x + 3)(5 x + 6)=0\). The solutions are \(\displaystyle x = - \frac{3}{4}\) and \(\displaystyle x = - \frac{6}{5}\)
\begin{question}Solve $- 20 x^{2} - 39 x - 18=0$
\soln{9cm}{Since the lead coefficient is negative factor out $-1$. This is an ac method factoring. The factors of $ac = 360$ that add up to $39$ are $24$ and $15$. Separating gives $(20 x^{2} + 24 x)+(15 x + 18)=0$. This gives $-1(4 x + 3)(5 x + 6)=0$. The solutions are $x = - \frac{3}{4}$ and $x = - \frac{6}{5}$}
\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 - 20 x^{2} - 39 x - 18=0 " src="/equation_images/%20%5Cdisplaystyle%20-%2020%20x%5E%7B2%7D%20-%2039%20x%20-%2018%3D0%20" alt="LaTeX: \displaystyle - 20 x^{2} - 39 x - 18=0 " data-equation-content=" \displaystyle - 20 x^{2} - 39 x - 18=0 " /> </p> </p><p> <p>Since the lead coefficient is negative factor out <img class="equation_image" title=" \displaystyle -1 " src="/equation_images/%20%5Cdisplaystyle%20-1%20" alt="LaTeX: \displaystyle -1 " data-equation-content=" \displaystyle -1 " /> . This is an ac method factoring. The factors of <img class="equation_image" title=" \displaystyle ac = 360 " src="/equation_images/%20%5Cdisplaystyle%20ac%20%3D%20360%20" alt="LaTeX: \displaystyle ac = 360 " data-equation-content=" \displaystyle ac = 360 " /> that add up to <img class="equation_image" title=" \displaystyle 39 " src="/equation_images/%20%5Cdisplaystyle%2039%20" alt="LaTeX: \displaystyle 39 " data-equation-content=" \displaystyle 39 " /> are <img class="equation_image" title=" \displaystyle 24 " src="/equation_images/%20%5Cdisplaystyle%2024%20" alt="LaTeX: \displaystyle 24 " data-equation-content=" \displaystyle 24 " /> and <img class="equation_image" title=" \displaystyle 15 " src="/equation_images/%20%5Cdisplaystyle%2015%20" alt="LaTeX: \displaystyle 15 " data-equation-content=" \displaystyle 15 " /> . Separating gives <img class="equation_image" title=" \displaystyle (20 x^{2} + 24 x)+(15 x + 18)=0 " src="/equation_images/%20%5Cdisplaystyle%20%2820%20x%5E%7B2%7D%20%2B%2024%20x%29%2B%2815%20x%20%2B%2018%29%3D0%20" alt="LaTeX: \displaystyle (20 x^{2} + 24 x)+(15 x + 18)=0 " data-equation-content=" \displaystyle (20 x^{2} + 24 x)+(15 x + 18)=0 " /> . This gives <img class="equation_image" title=" \displaystyle -1(4 x + 3)(5 x + 6)=0 " src="/equation_images/%20%5Cdisplaystyle%20-1%284%20x%20%2B%203%29%285%20x%20%2B%206%29%3D0%20" alt="LaTeX: \displaystyle -1(4 x + 3)(5 x + 6)=0 " data-equation-content=" \displaystyle -1(4 x + 3)(5 x + 6)=0 " /> . The solutions are <img class="equation_image" title=" \displaystyle x = - \frac{3}{4} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-%20%5Cfrac%7B3%7D%7B4%7D%20" alt="LaTeX: \displaystyle x = - \frac{3}{4} " data-equation-content=" \displaystyle x = - \frac{3}{4} " /> and <img class="equation_image" title=" \displaystyle x = - \frac{6}{5} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-%20%5Cfrac%7B6%7D%7B5%7D%20" alt="LaTeX: \displaystyle x = - \frac{6}{5} " data-equation-content=" \displaystyle x = - \frac{6}{5} " /> </p> </p>