If \(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \), find the values of x.
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Correct Answer: Option D
Explanation:
\(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \implies (1+2x)(3-x) - (-6) = -3\)
\(3 - x + 6x - 2x^{2} + 6 = -3\)
\(-2x^{2} + 5x + 3 + 6 + 3 = 0\)
Multiplying through with -1,
\(2x^{2} - 5x -12 = 0\)
\((2x + 3)(x - 4) = 0 \implies x = \frac{-3}{2} , 4\)
\(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \implies (1+2x)(3-x) - (-6) = -3\)
\(3 - x + 6x - 2x^{2} + 6 = -3\)
\(-2x^{2} + 5x + 3 + 6 + 3 = 0\)
Multiplying through with -1,
\(2x^{2} - 5x -12 = 0\)
\((2x + 3)(x - 4) = 0 \implies x = \frac{-3}{2} , 4\)