(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.