In reply to Robert Durran:
This is a similar question from a few years back. It uses compound shapes instead of conditional probability but the principles of deriving a quadratic from initial conditions are the same. Sorry some of the formatting isn't right and the diagram wouldn't copy, but it's basically a compound shape with sides in x. The important thing is the way the question is structured and the examiners comments.
1. The diagram below shows a 6-sided shape.
All the corners are right angles.
All measurements are given in centimetres.
The area of the shape is 25 cm2.
(a) Show that 6x2 + 17x ¡V 39 = 0
(3)
(b) (i) Solve the equation
6x2 + 17x ¡V 39 = 0
x = ¡K¡K¡K¡K¡K or x = ¡K¡K¡K¡K¡K
(ii) Hence work out the length of the longest side of the shape.
¡K¡K¡K¡K¡K¡K..cm
(4)
(Total 7 marks)
MS. (a) 6x2 + 11x ¡V 10 + 6x ¡V4 = 25
6x2 + 17x ¡V 39 = 0 3
M1 for an expression for the area involving either
(3x ¡V 2)(2x + 5) +2(3x ¡V 2)
or 3x(3x ¡V 2) + (3x ¡V 2)(7 ¡V x)
or 3x(2x + 5) ¡V 2(7 ¡V x)
or (3x ¡V 2)2 + 2(3x ¡V 2) + (3x ¡V 2)(7 ¡V x)
where in each case at least one of 2 or 3 product terms
must be correct
M1 (indep) for one correct expansion involving x2
A1 for simplification to final answer
3. This proved to be the first really challenging question for the candidates. "There is still a minority of candidates who do not understand that in part (a) they are required to derive the quadratic equation from given information. They give themselves away by trying to solve the equation as their answer to part (a)." The most commonly successful approach was to identify two rectangles of areas (3x ¡V 2)(2x +5) and 2(3x ¡V 2) respectively and then set the algebraic sum equal to 25. Further marks were then gained by using correct algebra to get to the given equation. Splitting the shape horizontally proved to be less successful as often the top rectangle was given the measurements (3x ¡V 2) and (2x + 5). Other methods involved splitting into the sum of three parts and working on the difference between the area of the full rectangle (2x + 5) by 3x and the small rectangle 2 by (2x + 5) ¡V (3x ¡V 2) although in many cases the second term was not worked out correctly.
Part (b) was generally tackled by using the formula. The usual error of not spotting that 172 ¡V 4 ¡Ñ 6 ¡Ñ (¡V 39) = 289 ¡V 936 is incorrect was often seen. Other errors included a faulty evaluation of as ¡V 17 b 35 ¡Ò 12 and 2 in the denominator rather than 12. Sometimes the negative sign was omitted from the second solution.
Some candidates realised that they could factorise the left hand side and often did so successfully. A minority once they had found the solutions reversed the signs.
The relevance to this thread is the way the question asks students to "show that" in exactly the same way as this question, and the examiners initial comments which clearly indicate that an approach based on solving the quadratic and subbing in is not acceptable for reasons I've given several times previously in this thread.
Post edited at 12:50