To solve the given equation, we need to evaluate and compare the determinants on both sides:
Given:
\(\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & -2 \\ 7 & 3\end{vmatrix}\)
The determinant of a 2x2 matrix \(\begin{vmatrix}a & b \\ c & d\end{vmatrix}\) is calculated as \(ad-bc\).
Let's calculate the determinant on the left side:
\(\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = (2x)(x) - (5)(8)\)
\(= 2x^2 - 40\)
Now, let's calculate the determinant on the right side:
\(\begin{vmatrix}6 & -2 \\ 7 & 3\end{vmatrix} = (6)(3) - (-2)(7)\)
\(= 18 + 14\)
\(= 32\)
Equating the determinants:
\(2x^2 - 40 = 32\)
Solving for \(x\), we get:
\(2x^2 = 72\)
\(x^2 = 36\)
Taking the square root on both sides, we find:
\(x = \pm 6\)
Thus, the values of \(x\) that satisfy the equation are \(\pm 6\). Hence, the correct answer is \(\pm 6\).