(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.
Explanation
(a) \(2Ax - Kx^{2}\) When x = 1, \(2A(1) - K(1^{2}) = 7\) \(2A - K = 7 ... (1)\) When x = 2, \(2A(2) - K(2^{2}) = 4\) \(4A - 4K = 4 \implies A - K = 1 ... (2)\) From (2), \(A = 1 + K\) (1) becomes : \(2(1 + K) - K = 7\) \(2 + 2K - K = 7 \implies 2 + K = 7\) \(\therefore K = 5\) \(A = 1 + K = 1 + 5 = 6\) \(A, K = 6, 5\) (b) \(x^{2} - 3x - 1 = 0\) \(a = 1, b = -3, c = -1\) Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-3)}}{2(1)}\) \(x = \frac{3 \pm \sqrt{9 + 12}}{2}\) \(x = \frac{3 \pm \sqrt{21}}{2} = \frac{3 \pm 4.583}{2}\) \(x = \frac{3 + 4.583}{2} ; x = \frac{3 - 4.583}{2}\) \(x = \frac{7.583}{2} = 3.7915 \approxeq 3.8\) (to 1 d.p) \(x = \frac{-1.583}{2} = -0.7915 \approxeq -0.8\) (to 1 d.p)