(a) Using the method of completing the square, solve, correct to 2 decimal places, \(\frac{x - 2}{4} = \frac{x + 2}{2x}\).
(b) 
In the diagram, PQRST is a circle with centre O. If PS is a diameter, RS//QT, and < QTS = 52°, find :
(i) < SQT ; (ii) < PQT.
(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° - 52° = 128° (opp. angles of a cyclic quadrilateral).
\(x_{1} = x_{2}\) (base angles of isosceles triangle)
\(x_{1} + x_{2} + 128° = 180°\) (sum of angles of a triangle)
\(2x_{2} = 180° - 128° = 52°\)
\(x_{2} = 26°\)
Hence, < SQT = 26°.
(ii) < PQS = 90° (angles in a semi-circle)
Hence, < PQT = 90° - 26° = 64°.
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