(a) Given that \(p = x + ym^{3}\), find m in terms of p, x and y. (b) Using the method of completing the square, find the roots of the equation \(x^{2} - 6x + 7 = 0\), correct to 1 decimal place. (c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.
Explanation
(a) \(p = x + ym^{3}\) \(ym^{3} = p - x\) \(m^{3} = \frac{p - x}{y}\) \(m = \sqrt[3]{\frac{p - x}{y}}\) (b) \(x^{2} - 6x + 7 = 0\) \(\implies x^{2} - 6x = -7\) \(x^{2} - 6x + (-3)^{2} = -7 + (-3)^{2}\) \(x^{2} - 6x + 9 = 2\) \((x - 3)^{2} = 2 \implies x - 3 = \pm \sqrt{2}\) \(x = 3 \pm \sqrt{2} = 3 \pm 1.414\) \(x = 4.414 ; 1.586\) \(x \approxeq 4.4 ; 1.6\) (c) Let the first odd number be z. The next positive odd number = z + 2 \(\therefore z(z + 2) = 195\) \(z^{2} + 2z = 195\) \(z^{2} + 2z - 195 = 0 \implies z^{2} - 13z + 15z - 195 = 0\) \(z(z - 13) + 15(z - 13) = 0 \implies \text{z = 13 or -15}\) Since we are told the odd numbers are positive, then z = 13 is the suitable answer. Hence the second odd number = 13 + 2 = 15. The two odd numbers are 13 and 15.