(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.
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°.