If \(\begin{vmatrix}2x&5\\8&x\end{vmatrix}=\begin{vmatrix}3&0\\4&-8\end{vmatrix}\) then value(s) of \(x\) is/are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\pm 2\sqrt{2}\)
Expanding each 2 × 2 determinant as \(ad - bc\):
Left side: \((2x)(x) - (5)(8) = 2x^2 - 40\)
Right side: \((3)(-8) - (0)(4) = -24\)
Equating the two:
\(2x^2 - 40 = -24\)
\(2x^2 = 16\)
\(x^2 = 8\)
A square equal to 8 has two square roots:
\(x = \pm 2\sqrt{2}\)
Why the other options are incorrect
Option (1) 3 — Incorrect. With \(x = 3\), the left side is \(18 - 40 = -22\), not \(-24\).
Option (2) \(2\sqrt{2}\) — Incorrect. It gives only one of the two values; \(x = -2\sqrt{2}\) also satisfies \(x^2 = 8\).
Option (4) 8 — Incorrect. 8 is the value of \(x^2\), not of x.