Question 6 Solve for z using the square root property. \[ \begin{array}{l} 6 z^{2}+68=1+5 \\ z=\frac{1}{\frac{1}{1}} \end{array} \] Entry Tipc Do not use a decimal approximation for square roots. To enter a number like \( 5 \sqrt{7} \), type 5 syri( (7). Rreview your answer before submitting? Furthere Tips wite your ancwers as a list, weth your ancwers separated by (a) comma(s). For example, if you get 4 and \( -\frac{2}{3} \) as your answers, then enter \( 4,-2 / 3 \) in the bou. Next Question

First, simplify the equation first by isolating the variable term. Subtract 1 and 5 from both sides to get \(6z^2 + 68 - 1 - 5 = 0\), leading to \(6z^2 + 62 = 0\). Now, rearrange this into \(6z^2 = -62\) and divide by 6 to obtain \(z^2 = -\frac{62}{6}\), which can be simplified to \(z^2 = -\frac{31}{3}\). Now apply the square root property! It's crucial to note that \(z\) has complex solutions since we're looking at a negative value under the square root. Therefore, we'll have \(z = \pm i\sqrt{\frac{31}{3}}\). Simplifying further gives us the final answers: \(z = \pm \frac{i \sqrt{93}}{3}\). So, your results are: \[ \frac{i \sqrt{93}}{3}, -\frac{i \sqrt{93}}{3} \]