Find the possible values of k for which the equation kx² + (k+5)x-4 = 0 has real roots.
- Find the possible values of k for which the equation kx² + (k+5)x-4 = 0 has real roots.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [M1] Uses b²-4ac oe: (k+5)² - 4k(-4) [* 0, where * could be = or any inequality sign]
- [A1] Forms a correct 3-term expression: k² + 26k +25
- [M1] Factorises k² + 26k + 25 or solves k² + 26k + 25 = 0 oe dep on first M1, FT their 3-term quadratic in k
- [A1] Correct critical values -1, - 25 soi
- [A1] k≤-25, k≥-1 mark final answer
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