(a) \(\frac{x - 2}{4} = \frac{x + 2}{2x}\) \(2x(x - 2) = 4(x + 2)\) \(2x^{2} - 4x = 4x + 8\) \(2x^{2} - 4x - 4x - 8 = 2x^{2} - 8x - 8 = 0\) Divide through by 2, we have \(x^{2} - 4x - 4 = 0\) \(x^{2} - 4x = 4\) Taking the square of \(\frac{b}{2}\) and add to both sides, \(x^{2} - 4x + (-2)^{2} = 4 + (-2)^{2}\) \((x - 2)^{2} = 8\) \(x - 2 = \pm {\sqrt{8}}\) \(x = 2 \pm \sqrt{8}\) \(x = 2 \pm 2.828\) \(x = 4.828\) or \(x = -0.828\). Hence, x = 4.83 or -0.83 (2 decimal place). (b)(i) In the diagram above, < QRS = 180^\circ - 52^\circ = 128^\circ (opp. angles of a cyclic quadrilateral). \(x_{1} = x_{2}\) (base angles of isosceles triangle) \(x_{1} + x_{2} + 128^\circ = 180^\circ\) (sum of angles of a triangle) \(2x_{2} = 180^\circ - 128^\circ = 52^\circ\) \(x_{2} = 26^\circ\) Hence, < SQT = 26^\circ. (ii) < PQS = 90^\circ (angles in a semi-circle) Hence, < PQT = 90^\circ - 26^\circ = 64^\circ.