(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.