\begin{tabular}{l} Question \\ Solve by completing the square: \( 5 x^{2}-27 x+4=0 \) \\ Select the correct answer below: \\ - \( \frac{27+\sqrt{649}}{10} \) and \( \frac{27-\sqrt{649}}{10} \) \\ \( \frac{-27+\sqrt{649}}{10} \) and \( \frac{-27-\sqrt{649}}{10} \) \\ . \\ -27+ \( \sqrt{649} \) and \( 27-\sqrt{649} \). \\ \hline\end{tabular}

To solve the equation \( 5x^{2}-27x+4=0 \) by completing the square, first divide everything by 5 to simplify things: \[ x^{2}-\frac{27}{5}x+\frac{4}{5}=0 \] Then, move the constant term to the other side: \[ x^{2}-\frac{27}{5}x=\frac{-4}{5} \] Next, complete the square on the left side. Take half of \(-\frac{27}{5}\), square it, and add it to both sides: \[ \left(-\frac{27}{10}\right)^{2}=\frac{729}{100} \] So, adding to both sides gives: \[ x^{2}-\frac{27}{5}x+\frac{729}{100}=\frac{729}{100}-\frac{4}{5} \] Now simplify the right side: \[ \frac{4}{5} = \frac{80}{100} \] Thus: \[ \frac{729}{100} - \frac{80}{100} = \frac{649}{100} \] Now you have: \[ \left(x-\frac{27}{10}\right)^{2}=\frac{649}{100} \] Taking the square root of both sides leads to: \[ x-\frac{27}{10}=\pm \frac{\sqrt{649}}{10} \] Finally, solving for x gives: \[ x=\frac{27+\sqrt{649}}{10} \quad \text{and} \quad x=\frac{27-\sqrt{649}}{10} \] Historical enthusiasts will be pleased to know that completing the square dates back to ancient Babylon! They solved quadratic equations using geometric methods, remarkably similar to our algebraic approach today. For practical applications, understanding the method of completing the square is pivotal in fields like physics and engineering. It's not just an abstract concept but a technique that allows for finding maximums and minimums in functions, essential for optimization problems in real-world scenarios!