16N.1.AHL.TZ0.H_5

pestleMathematicsAAHLPaper 116N· sl-2-7-solutions-of-quadratic-equations-and-inequalities-discriminant-and-nature-of-rootssource ↗

The quadratic equation ${x^2} - 2kx + (k - 1) = 0$ has roots $\alpha$ and $\beta$ such that ${\alpha ^2} + {\beta ^2} = 4$ . Without solving the equation, find the possible values of the real number $k$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\alpha + \beta = 2k$ A1

$\alpha \beta = k - 1$ A1

${(\alpha + \beta )^2} = 4{k^2} \Rightarrow {\alpha ^2} + {\beta ^2} + 2\underbrace {\alpha \beta }_{k - 1} = 4{k^2}$ (M1)

${\alpha ^2} + {\beta ^2} = 4{k^2} - 2k + 2$

${\alpha ^2} + {\beta ^2} = 4 \Rightarrow 4{k^2} - 2k - 2 = 0$ A1

attempt to solve quadratic (M1)

$k = 1,{\text{ }} - \frac{1}{2}$ A1

[6 marks]

Examiners’ report

[N/A]