| Date | May 2009 | Marks available | 5 | Reference code | 09M.2.sl.TZ2.7 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The quadratic equation \(k{x^2} + (k - 3)x + 1 = 0\) has two equal real roots.
Find the possible values of k.
Write down the values of k for which \({x^2} + (k - 3)x + k = 0\) has two equal real roots.
Markscheme
attempt to use discriminant (M1)
correct substitution, \({(k - 3)^2} - 4 \times k \times 1\) (A1)
setting their discriminant equal to zero M1
e.g. \({(k - 3)^2} - 4 \times k \times 1 = 0\) , \({k^2} - 10k + 9 = 0\)
\(k = 1\) , \(k = 9\) A1A1 N3
[5 marks]
\(k = 1\) , \(k = 9\) A2 N2
[2 marks]
Examiners report
Although some candidates correctly considered the discriminant to find the possible values of k , many of them did not set it equal to \(0\), writing an inequality instead.
In part (b), some students realized that the discriminants in parts (a) and (b) were the same, earning follow through marks just by writing the same (often incorrect) answers they got in part (a). Many, however, did not see the connection between the two parts.
