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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle - 10 x^{2} + 3 x + 18=0\)
Since the lead coefficient is negative factor out \(\displaystyle -1\). This is an ac method factoring. The factors of \(\displaystyle ac = -180\) that add up to \(\displaystyle -3\) are \(\displaystyle -15\) and \(\displaystyle 12\). Separating gives \(\displaystyle (10 x^{2} - 15 x)+(12 x - 18)=0\). This gives \(\displaystyle -1(5 x + 6)(2 x - 3)=0\). The solutions are \(\displaystyle x = - \frac{6}{5}\) and \(\displaystyle x = \frac{3}{2}\)
\begin{question}Solve $- 10 x^{2} + 3 x + 18=0$
\soln{9cm}{Since the lead coefficient is negative factor out $-1$. This is an ac method factoring. The factors of $ac = -180$ that add up to $-3$ are $-15$ and $12$. Separating gives $(10 x^{2} - 15 x)+(12 x - 18)=0$. This gives $-1(5 x + 6)(2 x - 3)=0$. The solutions are $x = - \frac{6}{5}$ and $x = \frac{3}{2}$}
\end{question}
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\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Solve <img class="equation_image" title=" \displaystyle - 10 x^{2} + 3 x + 18=0 " src="/equation_images/%20%5Cdisplaystyle%20-%2010%20x%5E%7B2%7D%20%2B%203%20x%20%2B%2018%3D0%20" alt="LaTeX: \displaystyle - 10 x^{2} + 3 x + 18=0 " data-equation-content=" \displaystyle - 10 x^{2} + 3 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 = -180 " src="/equation_images/%20%5Cdisplaystyle%20ac%20%3D%20-180%20" alt="LaTeX: \displaystyle ac = -180 " data-equation-content=" \displaystyle ac = -180 " /> that add up to <img class="equation_image" title=" \displaystyle -3 " src="/equation_images/%20%5Cdisplaystyle%20-3%20" alt="LaTeX: \displaystyle -3 " data-equation-content=" \displaystyle -3 " /> are <img class="equation_image" title=" \displaystyle -15 " src="/equation_images/%20%5Cdisplaystyle%20-15%20" alt="LaTeX: \displaystyle -15 " data-equation-content=" \displaystyle -15 " /> and <img class="equation_image" title=" \displaystyle 12 " src="/equation_images/%20%5Cdisplaystyle%2012%20" alt="LaTeX: \displaystyle 12 " data-equation-content=" \displaystyle 12 " /> . Separating gives <img class="equation_image" title=" \displaystyle (10 x^{2} - 15 x)+(12 x - 18)=0 " src="/equation_images/%20%5Cdisplaystyle%20%2810%20x%5E%7B2%7D%20-%2015%20x%29%2B%2812%20x%20-%2018%29%3D0%20" alt="LaTeX: \displaystyle (10 x^{2} - 15 x)+(12 x - 18)=0 " data-equation-content=" \displaystyle (10 x^{2} - 15 x)+(12 x - 18)=0 " /> . This gives <img class="equation_image" title=" \displaystyle -1(5 x + 6)(2 x - 3)=0 " src="/equation_images/%20%5Cdisplaystyle%20-1%285%20x%20%2B%206%29%282%20x%20-%203%29%3D0%20" alt="LaTeX: \displaystyle -1(5 x + 6)(2 x - 3)=0 " data-equation-content=" \displaystyle -1(5 x + 6)(2 x - 3)=0 " /> . The solutions are <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} " /> and <img class="equation_image" title=" \displaystyle x = \frac{3}{2} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B3%7D%7B2%7D%20" alt="LaTeX: \displaystyle x = \frac{3}{2} " data-equation-content=" \displaystyle x = \frac{3}{2} " /> </p> </p>