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If ∣2x58x∣=∣304−8∣\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 3 & 0 \\ 4 & -8 \end{vmatrix}, then value(s) of x is/are

Solution

✅ Correct Option: 3

For a 2×2 matrix ∣abcd∣\begin{vmatrix} a & b \\ c & d \end{vmatrix}, the determinant equals (a×d)−(b×c)(a × d) - (b × c).


Calculate the left side determinant:

∣2x58x∣\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}

=(2x)(x)−(5)(8)= (2x)(x) - (5)(8)

=2x2−40= 2x^2 - 40


Calculate the right side determinant:

∣304−8∣\begin{vmatrix} 3 & 0 \\ 4 & -8 \end{vmatrix}

=(3)(−8)−(0)(4)= (3)(-8) - (0)(4)

=−24= -24


Set the determinants equal:

2x2−40=−242x^2 - 40 = -24

2x2=162x^2 = 16

x2=8x^2 = 8

x=±8x = \pm\sqrt{8}

x=±22x = \pm 2\sqrt{2}


Therefore, the values of xx are ±22\pm 2\sqrt{2}.

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