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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle - 8 x^{2} - 26 x - 15=0\)
Since the lead coefficient is negative factor out \(\displaystyle -1\). This is an ac method factoring. The factors of \(\displaystyle ac = 120\) that add up to \(\displaystyle 26\) are \(\displaystyle 6\) and \(\displaystyle 20\). Separating gives \(\displaystyle (8 x^{2} + 6 x)+(20 x + 15)=0\). This gives \(\displaystyle -1(2 x + 5)(4 x + 3)=0\). The solutions are \(\displaystyle x = - \frac{5}{2}\) and \(\displaystyle x = - \frac{3}{4}\)
\begin{question}Solve $- 8 x^{2} - 26 x - 15=0$
\soln{9cm}{Since the lead coefficient is negative factor out $-1$. This is an ac method factoring. The factors of $ac = 120$ that add up to $26$ are $6$ and $20$. Separating gives $(8 x^{2} + 6 x)+(20 x + 15)=0$. This gives $-1(2 x + 5)(4 x + 3)=0$. The solutions are $x = - \frac{5}{2}$ and $x = - \frac{3}{4}$}
\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 - 8 x^{2} - 26 x - 15=0 " src="/equation_images/%20%5Cdisplaystyle%20-%208%20x%5E%7B2%7D%20-%2026%20x%20-%2015%3D0%20" alt="LaTeX: \displaystyle - 8 x^{2} - 26 x - 15=0 " data-equation-content=" \displaystyle - 8 x^{2} - 26 x - 15=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 = 120 " src="/equation_images/%20%5Cdisplaystyle%20ac%20%3D%20120%20" alt="LaTeX: \displaystyle ac = 120 " data-equation-content=" \displaystyle ac = 120 " /> that add up to <img class="equation_image" title=" \displaystyle 26 " src="/equation_images/%20%5Cdisplaystyle%2026%20" alt="LaTeX: \displaystyle 26 " data-equation-content=" \displaystyle 26 " /> are <img class="equation_image" title=" \displaystyle 6 " src="/equation_images/%20%5Cdisplaystyle%206%20" alt="LaTeX: \displaystyle 6 " data-equation-content=" \displaystyle 6 " /> and <img class="equation_image" title=" \displaystyle 20 " src="/equation_images/%20%5Cdisplaystyle%2020%20" alt="LaTeX: \displaystyle 20 " data-equation-content=" \displaystyle 20 " /> . Separating gives <img class="equation_image" title=" \displaystyle (8 x^{2} + 6 x)+(20 x + 15)=0 " src="/equation_images/%20%5Cdisplaystyle%20%288%20x%5E%7B2%7D%20%2B%206%20x%29%2B%2820%20x%20%2B%2015%29%3D0%20" alt="LaTeX: \displaystyle (8 x^{2} + 6 x)+(20 x + 15)=0 " data-equation-content=" \displaystyle (8 x^{2} + 6 x)+(20 x + 15)=0 " /> . This gives <img class="equation_image" title=" \displaystyle -1(2 x + 5)(4 x + 3)=0 " src="/equation_images/%20%5Cdisplaystyle%20-1%282%20x%20%2B%205%29%284%20x%20%2B%203%29%3D0%20" alt="LaTeX: \displaystyle -1(2 x + 5)(4 x + 3)=0 " data-equation-content=" \displaystyle -1(2 x + 5)(4 x + 3)=0 " /> . The solutions are <img class="equation_image" title=" \displaystyle x = - \frac{5}{2} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-%20%5Cfrac%7B5%7D%7B2%7D%20" alt="LaTeX: \displaystyle x = - \frac{5}{2} " data-equation-content=" \displaystyle x = - \frac{5}{2} " /> and <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} " /> </p> </p>