(a) The value of the expression \(2Ax - Kx^{2}\) is 7 when x = 1 and 4 when x = 2. Find the values of the constants A and K.
(b) Solve the equation \(x^{2} - 3x - 1 = 0\), giving your answers correct to 1 decimal place.
(a) \(2Ax - Kx^{2}\)
When \(x = 1\),
\(2A(1) - K(1^{2}) = 7\)
\(2A - K = 7 \quad \ldots (1)\)
When \(x = 2\),
\(2A(2) - K(2^{2}) = 4\)
\(4A - 4K = 4\)
\(A - K = 1 \quad \ldots (2)\)
From (2),
\(A = 1 + K\)
Substitute into (1):
\(2(1 + K) - K = 7\)
\(2 + 2K - K = 7\)
\(K = 5\)
\(A = 1 + 5 = 6\)
\(\therefore A = 6,\quad K = 5\)
(b) \(x^{2} - 3x - 1 = 0\)
\(a = 1,\quad b = -3,\quad c = -1\)
Using the quadratic formula,
\[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]
\[x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-1)}}{2(1)}\]
\[x = \frac{3 \pm \sqrt{9 + 4}}{2}\]
\[x = \frac{3 \pm \sqrt{13}}{2}\]
\[x = \frac{3 \pm 3.606}{2}\]
\[x = \frac{3 + 3.606}{2} = 3.303 \approx 3.3\]
or
\[x = \frac{3 - 3.606}{2} = -0.303 \approx -0.3\]
\(\therefore x = 3.3\) or \(x = -0.3\), correct to 1 decimal place.