WAEC 2016 · Paper 2 · Q7

(a)
Using the completing the square method, solve, correct to 2 decimal places, $\dfrac{x - 2}{4} = \dfrac{x + 2}{2x}$ .
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $PQRST$ is a circle with centre $O$ . If $PS$ is a diameter, $RS \parallel QT$ , $|QR| = |RS|$ and $\angle QTS = 52^\circ$ , find: (i) $\angle SQT$ ; (ii) $\angle PQT$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Clear the fractions by multiplying both sides by $4 \times 2x$ : $2x(x - 2) = 4(x + 2)$ , so $2x^2 - 4x = 4x + 8$ .
Collect everything on one side and divide by 2: $x^2 - 4x - 4 = 0$ , so $x^2 - 4x = 4$ .
Complete the square by adding $\left(\frac{-4}{2}\right)^2 = 4$ to both sides: $x^2 - 4x + 4 = 8$ , so $(x - 2)^2 = 8$ .
Take square roots: $x - 2 = \pm\sqrt8 = \pm 2.828$ .
So $x = 4.83$ or $x = -0.83$ to 2 decimal places.

(b)(i)

$QRST$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$ : $\angle QRS = 180^\circ - 52^\circ$
$= 128^\circ$ .
$|QR| = |RS|$ , so triangle $QRS$ is isosceles and $\angle RSQ = \frac{180^\circ - 128^\circ}{2}$
$= 26^\circ$ .
$RS \parallel QT$ , so $\angle SQT = \angle RSQ = 26^\circ$ (alternate angles).

(ii)

$PS$ is a diameter, so $\angle PQS = 90^\circ$ (angle in a semicircle).
Then $\angle PQT = 90^\circ - 26^\circ$
$= 64^\circ$ .

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