(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)