A-Level Equations and inequalities: worked solution

Question

Using part (a) or otherwise, solve |x - a| > 1/|x - a| for a real constant a > 1.

Both sides depend only on |x - a|, so simplify by letting t = |x - a| Note t >= 0, and t is not 0, because 1/|x - a| must be defined (so x is not a). The inequality becomes t > 1/t Multiplying by t is safe here, because t > 0 and multiplying by a positive number does not reverse an inequality: t² > 1 t² - 1 > 0 (t - 1)(t + 1) > 0 Since t > 0, the factor t + 1 is always positive, so the product is positive exactly when t - 1 > 0: t > 1 This matches the graphical picture: y = t and y = 1/t meet where t² = 1, that is where (x - a)² = 1, giving the critical values x - a = 1 or x - a = -1 x = a + 1 or x = a - 1 and y = t lies above y = 1/t only beyond that crossing point, i.e. for t > 1. Finally undo the modulus. |x - a| > 1 means x - a is more than 1 away from 0 in either direction: x - a > 1 or x - a < -1 x > a + 1 or x < a - 1 Solution: x < a - 1 or x > a + 1. (Both critical values are positive here since a > 1, and x = a itself is automatically excluded because it lies between them.)