(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.
(b) Solve the equation \(x^{2} - 3x - 1 = 0\), giving your answers correct to 1 decimal place.
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Correct Answer: Option n
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)
(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)