User:Espilceranul



\text {2009 EOY Bonus Question 17}\,$$
 * $$\text {Make p the subject of the formula.}\,$$
 * $$z^bp^{25c} = \sqrt{2z^{3b}\cdot p^{25c} + z^{2a} - z^{4b}}\,$$
 * $$z^{2b}p^{50c} = 2z^3b{p^25c} + z^{2a} - z^{4b}\,$$
 * $$z^{2b}p^{50c} - 2z^{3b}p^{25c} + z^{4b} = z^{2a}\,$$
 * $$z^{2b}(p^{50c} - 2z^bp^{25c} + z^{2b}) = z^{2a}\,$$
 * $$z^{2b}(p^{25c} - z^b)^2 = z^{2a}\,$$
 * $$p^{25c} - z^b = \pm \frac {z^a}{z^b}\,$$
 * $$\therefore p = \sqrt[25c]{\frac {z^{2b} \pm z^a}{z^b}}\,$$
 * $$\text {they said assume c is an odd number so no plus minus on the last one}\,$$