Practicing Success
Practicing Success
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If $\begin{vmatrix} 2 & 4\\5 & 1\end{vmatrix}=\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}$ then, x is equal to : |
1 $2\sqrt{3}$ $±\sqrt{3}$ $±2\sqrt{3}$ |
$±\sqrt{3}$ |
The correct answer is Option (3) → $±\sqrt{3}$ $\begin{vmatrix} 2 & 4\\5 & 1\end{vmatrix}=\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}$ so $2-20=2x^2-24$ $2x^2=24-20+2$ $2x^2=6$ $x^2=3⇒x=±\sqrt{3}$ |
