(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 ci...

(a) Completing the square.

Clear the fractions in \(\dfrac{x-2}{4}=\dfrac{x+2}{2x}\) by cross-multiplying:

\[2x(x-2)=4(x+2)\]

\[2x^{2}-4x=4x+8\]

\[2x^{2}-8x-8=0\]

Divide through by 2 to make the coefficient of \(x^{2}\) equal to 1:

\[x^{2}-4x-4=0 \quad\Rightarrow\quad x^{2}-4x=4\]

Add the square of half the coefficient of \(x\), i.e. \(\left(\tfrac{-4}{2}\right)^{2}=4\), to both sides:

\[x^{2}-4x+4=4+4\]

\[(x-2)^{2}=8\]

\[x-2=\pm\sqrt{8}=\pm 2.8284\]

\[x=2+2.8284=4.83 \quad\text{or}\quad x=2-2.8284=-0.83\]

\[\boxed{x=4.83 \text{ or } x=-0.83}\ \text{(to 2 d.p.)}\]

(b) Circle PQRST, centre O.

From the diagram: PS is a diameter, the tick marks show \(QR=RS\), the arrows show \(RS\parallel QT\), and \(\angle QTS=52^\circ\).

(i) \(\angle SQT\).

\(\angle QTS=52^\circ\) is an angle at the circumference standing on the arc \(QRS\), so

\[\text{arc } QR+\text{arc } RS=2\times 52^\circ=104^\circ.\]

Since \(QR=RS\), the equal chords cut off equal arcs, so \(\text{arc } QR=\text{arc } RS=52^\circ\).

Because \(RS\parallel QT\), the arcs between the parallel chords are equal, giving \(\text{arc } ST=\text{arc } QR=52^\circ\).

\(\angle SQT\) stands at the circumference on arc \(ST\):

\[\angle SQT=\tfrac{1}{2}\,\text{arc } ST=\tfrac{1}{2}\times 52^\circ=\boxed{26^\circ}.\]

(ii) \(\angle PQT\).

PS is a diameter, so \(\angle PQS=90^\circ\) (angle in a semicircle).

The ray \(QT\) lies between \(QP\) and \(QS\), so

\[\angle PQT=\angle PQS-\angle SQT=90^\circ-26^\circ=\boxed{64^\circ}.\]