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If ∣2x28x∣=∣−4−24−2∣\begin{vmatrix} 2x & 2 \\ 8 & x \end{vmatrix} = \begin{vmatrix} -4 & -2 \\ 4 & -2 \end{vmatrix} then x=x =

Solution

✅ Correct Option: 2

LHS =2x2−16=2x^2-16 and RHS =8+8=16=8+8=16.

So 2x2=322x^2=32, x2=16x^2=16 and x=±4x=\pm 4.

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