Find the values of k such that \(f(x) = 0\) has two equal roots.

Each value of k is equally likely for \( - 5 \le k \le 5\) . Find the probability that \(f(x) = 0\) has no roots.

METHOD 1

evidence of discriminant     (M1)

e.g. \({b^2} - 4ac\) , discriminant = 0

correct substitution into discriminant     A1

e.g. \({k^2} - 4 \times \frac{1}{2} \times 8\) , \({k^2} - 16 = 0\)

\(k = \pm 4\)     A1A1     N3

METHOD 2

recognizing that equal roots means perfect square     (R1)

e.g. attempt to complete the square, \(\frac{1}{2}({x^2} + 2kx + 16)\)

correct working

e.g. \(\frac{1}{2}{(x + k)^2}\) ,  \(\frac{1}{2}{k^2} = 8\)     A1

\(k = \pm 4\)     A1A1     N3

[4 marks]

evidence of appropriate approach     (M1)

e.g. \({b^2} - 4ac < 0\)

correct working for k     A1

e.g. \( - 4 < k < 4\) , \({k^2} < 16\) , list all correct values of k

\(p = \frac{7}{{11}}\)     A2     N3

[4 marks]

A good number of candidates were successful in using the discriminant to find the correct values of k in part (a), however, there were many who tried to use the quadratic formula without recognizing the significance of the discriminant.

Part (b) was very poorly done by nearly all candidates. Common errors included finding the wrong values for k, and not realizing that there were 11 possible values for k.