Solve the equation by completing the square. \( x^{2}-\frac{5}{2} x-\frac{39}{16}=0 \) Determine the appropriate process for rearranging the equation before completing the square to obtain an equation of the form \( (x+k)^{2}= \) d. Select the correct choice below and fill in the answer box(es) within your choice. (Type an integer or a simplified fraction.) A. \( x^{2}-\frac{5}{2} x-\frac{39}{16}=0 \Rightarrow x^{2}-\frac{5}{2} x=\frac{39}{16} \Rightarrow x^{2}-\frac{5}{2} x+\frac{25}{16}=\frac{39}{16}+\frac{25}{16} \) B. \( x^{2}-\frac{5}{2} x-\frac{39}{16}=0 \Rightarrow x^{2}-\frac{5}{2} x=\frac{39}{16} \Rightarrow x^{2}-\frac{5}{2} x+\quad=\frac{39}{16} \) C. \( x^{2}-\frac{5}{2} x-\frac{39}{16}=0 \Rightarrow \frac{2}{5} x^{2}-x=\frac{39}{40} \Rightarrow \frac{2}{5} x^{2}-x+=\frac{39}{40}+ \) D. \( x^{2}-\frac{5}{2} x-\frac{39}{16}=0 \Rightarrow \frac{2}{5} x^{2}-x=\frac{39}{40} \Rightarrow \frac{2}{5} x^{2}-x+\quad=\frac{39}{40} \) The solution set is \( \left\{\begin{array}{l}\text { A. } \\ \text { (Simplify your answer. Type an exact answer, using radicals as needed. Use integers or simplified fractions for any } \\ \text { numbers in the expression. Use a comma to separate answers as needed.) }\end{array}\right. \) \( \begin{array}{l}\text { a }\end{array} \)
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The Deep Dive
To solve this equation by completing the square, we need to first isolate the \(x^2\) terms on one side. The equation starts as \(x^{2} - \frac{5}{2} x - \frac{39}{16} = 0\). Choosing option A is the way to go—this option correctly rearranges the equation and prepares for the completion of the square. Once we add \(\frac{25}{16}\) to both sides, we convert our equation to \((x - \frac{5}{4})^2 = \text{something}\). After completing the square and simplifying, the next step is to square root both sides, leading to two potential solutions for \(x\). Don’t forget to consider both the positive and negative square roots! Now, once you finish solving the equation using your selected path, double-check the found solutions by substituting them back into the original equation. This ensures no calculation errors crept in during the process!