(a) Using the method of completing the square, solve, correct to 2 decimal places, \(\frac{x...

Explanation

(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°.