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Questions: Algebra BusinessCalculus
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Solve \(\displaystyle - 15 x^{2} + 52 x - 45=0\)
Since the lead coefficient is negative factor out \(\displaystyle -1\). This is an ac method factoring. The factors of \(\displaystyle ac = 675\) that add up to \(\displaystyle -52\) are \(\displaystyle -27\) and \(\displaystyle -25\). Separating gives \(\displaystyle (15 x^{2} - 27 x)+(45 - 25 x)=0\). This gives \(\displaystyle -1(3 x - 5)(5 x - 9)=0\). The solutions are \(\displaystyle x = \frac{5}{3}\) and \(\displaystyle x = \frac{9}{5}\)
\begin{question}Solve $- 15 x^{2} + 52 x - 45=0$
\soln{9cm}{Since the lead coefficient is negative factor out $-1$. This is an ac method factoring. The factors of $ac = 675$ that add up to $-52$ are $-27$ and $-25$. Separating gives $(15 x^{2} - 27 x)+(45 - 25 x)=0$. This gives $-1(3 x - 5)(5 x - 9)=0$. The solutions are $x = \frac{5}{3}$ and $x = \frac{9}{5}$}
\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 - 15 x^{2} + 52 x - 45=0 " src="/equation_images/%20%5Cdisplaystyle%20-%2015%20x%5E%7B2%7D%20%2B%2052%20x%20-%2045%3D0%20" alt="LaTeX: \displaystyle - 15 x^{2} + 52 x - 45=0 " data-equation-content=" \displaystyle - 15 x^{2} + 52 x - 45=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 = 675 " src="/equation_images/%20%5Cdisplaystyle%20ac%20%3D%20675%20" alt="LaTeX: \displaystyle ac = 675 " data-equation-content=" \displaystyle ac = 675 " /> that add up to <img class="equation_image" title=" \displaystyle -52 " src="/equation_images/%20%5Cdisplaystyle%20-52%20" alt="LaTeX: \displaystyle -52 " data-equation-content=" \displaystyle -52 " /> are <img class="equation_image" title=" \displaystyle -27 " src="/equation_images/%20%5Cdisplaystyle%20-27%20" alt="LaTeX: \displaystyle -27 " data-equation-content=" \displaystyle -27 " /> and <img class="equation_image" title=" \displaystyle -25 " src="/equation_images/%20%5Cdisplaystyle%20-25%20" alt="LaTeX: \displaystyle -25 " data-equation-content=" \displaystyle -25 " /> . Separating gives <img class="equation_image" title=" \displaystyle (15 x^{2} - 27 x)+(45 - 25 x)=0 " src="/equation_images/%20%5Cdisplaystyle%20%2815%20x%5E%7B2%7D%20-%2027%20x%29%2B%2845%20-%2025%20x%29%3D0%20" alt="LaTeX: \displaystyle (15 x^{2} - 27 x)+(45 - 25 x)=0 " data-equation-content=" \displaystyle (15 x^{2} - 27 x)+(45 - 25 x)=0 " /> . This gives <img class="equation_image" title=" \displaystyle -1(3 x - 5)(5 x - 9)=0 " src="/equation_images/%20%5Cdisplaystyle%20-1%283%20x%20-%205%29%285%20x%20-%209%29%3D0%20" alt="LaTeX: \displaystyle -1(3 x - 5)(5 x - 9)=0 " data-equation-content=" \displaystyle -1(3 x - 5)(5 x - 9)=0 " /> . The solutions are <img class="equation_image" title=" \displaystyle x = \frac{5}{3} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B5%7D%7B3%7D%20" alt="LaTeX: \displaystyle x = \frac{5}{3} " data-equation-content=" \displaystyle x = \frac{5}{3} " /> and <img class="equation_image" title=" \displaystyle x = \frac{9}{5} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B9%7D%7B5%7D%20" alt="LaTeX: \displaystyle x = \frac{9}{5} " data-equation-content=" \displaystyle x = \frac{9}{5} " /> </p> </p>