If $\begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}3&0\\4&-8\end{vmatrix}$ then value(s) of $x$ is/are

$±2\sqrt{2}$

The correct answer is Option (3) → $±2\sqrt{2}$

Given:

$\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 3 & 0 \\ 4 & -8 \end{vmatrix}$

Compute determinants:

Left side: $(2x)(x) - (5)(8) = 2x^2 - 40$

Right side: $(3)(-8) - (0)(4) = -24$

Equating:

$2x^2 - 40 = -24$

$2x^2 = 16$

$x^2 = 8$

$x = \pm 2\sqrt{2}$

$x = \pm 2\sqrt{2}$