Consider the equation \({x^2} + (k - 1)x + 1 = 0\) , where k is a real number.

Find the values of k for which the equation has two equal real solutions.

METHOD 1

evidence of valid approach     (M1)

e.g. \({b^2} - 4ac\) , quadratic formula

correct substitution into \({b^2} - 4ac\) (may be seen in formula)     (A1)

e.g. \({(k - 1)^2} - 4 \times 1 \times 1\) , \({(k - 1)^2} - 4\) , \({k^2} - 2k - 3\)

setting their discriminant equal to zero     M1

e.g. \(\Delta  = 0,{(k - 1)^2} - 4 = 0\)

attempt to solve the quadratic     (M1)

e.g. \({(k - 1)^2} = 4\) , factorizing

correct working     A1

e.g. \((k - 1) = \pm 2\) , \((k - 3)(k + 1)\)

\(k = - 1\) , \(k = 3\) (do not accept inequalities)     A1A1     N2

[7 marks]

METHOD 2

recognizing perfect square     (M1)

e.g. \({(x + 1)^2} = 0\) , \({(x - 1)^2}\)

correct expansion     (A1)(A1)

e.g. \({x^2} + 2x + 1 = 0\) , \({x^2} - 2x + 1\)

equating coefficients of x     A1A1

e.g. \(k - 1 = - 2\) , \(k - 1 = 2\)

\(k = - 1\) , \(k = 3\)     A1A1     N2

[7 marks]

Most candidates approached this question correctly by using the discriminant, and many were successful in finding both of the required values of k. There did seem to be some confusion about the expression "two equal real solutions", as some candidates approached the question as though the equation had two distinct real roots, using \({b^2} - 4ac > 0\) , rather than \({b^2} - 4ac = 0\) .

There were also a good number who recognized that the quadratic must be a perfect square, although many who used this method found only one of the two possible values of k. In addition, there were many unsuccessful candidates who tried to use the entire quadratic formula as though they were solving for x, without ever seeming to realize the significance of the discriminant.