(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...
Question 1 Report
(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.
Answer Details
(a) \(p = x + ym^3\). Make \(m\) the subject.
\(ym^3 = p - x \ \Rightarrow\ m^3 = \dfrac{p-x}{y} \ \Rightarrow\ \mathbf{m = \sqrt[3]{\dfrac{p-x}{y}}}\)
(b) \(x^2 - 6x + 7 = 0\) by completing the square.
\(x^2 - 6x = -7\)
Add \(\left(\tfrac{6}{2}\right)^2 = 9\) to both sides:
\((x-3)^2 = -7 + 9 = 2\)
\(x - 3 = \pm\sqrt{2} = \pm 1.414\)
\(x = 3 + 1.414 = 4.4\quad\text{or}\quad x = 3 - 1.414 = 1.6\) (1 d.p.)
(c) Let the consecutive odd numbers be \(n\) and \(n+2\).
\(n(n+2) = 195 \ \Rightarrow\ n^2 + 2n - 195 = 0\)
\((n+15)(n-13) = 0 \ \Rightarrow\ n = -15\ \text{or}\ n = 13\)
Since the numbers are positive, \(n = 13\). The numbers are \(\mathbf{13}\) and \(\mathbf{15}\).
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