WAEC - Further Mathematics (2015 - No. 26)

If \(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \), find the values of x.

\(x = 3, -2\)

\(x = 4, \frac{-2}{3}\)

\(x = -4, \frac{3}{2}\)

\(x = 4, \frac{-3}{2}\)

\(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \implies (1+2x)(3-x) - (-6) = -3\)

\(3 - x + 6x - 2x^{2} + 6 = -3\)

\(-2x^{2} + 5x + 3 + 6 + 3 = 0\)

Multiplying through with -1,

\(2x^{2} - 5x -12 = 0\)

\((2x + 3)(x - 4) = 0 \implies x = \frac{-3}{2} , 4\)