Further Mathematics Paper 2, Nov/Dec. 2012
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 2

For what values of k are the root of the equation
(k+3) x2 + (6 – 2k)x + k – 1 = 0 real?

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Observation

This question was reported to be quite popular among the candidates and their performance was reported as fair.  The report stated that majority of the candidates could not simplify the resulting inequality correctly.  Others were reported to have worked with the sign of equality instead of inequality and they lost some marks.  Candidates were expected to recall that if the equation ax2 + bx + c = 0 has real roots, then b2 ≥ 4ac.  Comparing ax2 + bx + c = 0 with the given equation, (k +3) = a, (6-2k) = b and (k-1) = c.  Therefore, for the given equation to have real roots, (6-2k)2 ≥ 4(k+3)(k-1) by expanding both sides of the inequality sign and bringing like terms together, we obtain k ≤ .