If $A\begin{vmatrix}x-2&-3\\3x&2x\end{vmatrix}=3$. What are the possible valve for x?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-3, \frac{1}{2}\)
$A\begin{vmatrix}x-2&-3\\3x&2x\end{vmatrix}=3$
$⇒2x(x-2)+9x=3$
$⇒2x^2-4x+9x=3$
$⇒2x^2+5x-3=0$
$⇒2x^2+6x-x-3=0$
$⇒2x(x+3)-1(x+3)=0$
$⇒(x+3)(2x-1)=0$
$⇒x=-3\,or\,\frac{1}{2}$