If \( \begin{vmatrix}8 & -4\\5 & -2\end{vmatrix}=\begin{vmatrix}x & 6\\2 & 2x\end{vmatrix} \), then \( |x| \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \sqrt{8} \)
The left-hand determinant is
\( \displaystyle \begin{vmatrix}8 & -4\\5 & -2\end{vmatrix}=8(-2)-(-4)(5) \)
\( \displaystyle =-16+20=4 \)
The right-hand determinant is
\( \displaystyle \begin{vmatrix}x & 6\\2 & 2x\end{vmatrix}=2x^2-12 \)
Equating the two:
\( \displaystyle 2x^2-12=4 \)
\( \displaystyle x^2=8 \)
\( \displaystyle x=\pm\sqrt{8} \)
\( \displaystyle |x|=\sqrt{8} \)
Why the other options are incorrect
Option (1) 8 — Incorrect. 8 is the value of \( x^2 \), not of |x|.
Option (2) -8 — Incorrect. An absolute value can never be negative.
Option (4) 2 — Incorrect. With |x| = 2, \( 2x^2-12=-4 \), not 4.