Higher June 2024 Paper 2R Q20
20 Solve \(\;6x^2 - 7x - 20 \gt 0\)
Show clear algebraic working.
(4)
| Scheme | Marks |
|---|---|
eg \((3x + 4)(2x - 5)\) or \((x =)\, \dfrac{--7 \pm \sqrt{(-7)^2 - 4 \times 6 \times (-20)}}{2 \times 6}\) oe or \(6\left[\left(x - \dfrac{7}{12}\right)^2 - \left(\dfrac{7}{12}\right)^2\right] - 20\) oe | M1 |
| \((x =)\, -\dfrac{4}{3}\) and \(\dfrac{5}{2}\) oe | A1 |
| M1ft | |
Working required Answer: \(x \lt -\dfrac{4}{3}\) \(x \gt \dfrac{5}{2}\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: first step to finding the critical values - if factorising (in the form \((ax + b)\) where \(a\) and \(b\) are integers), allow brackets which expanded give 2 out of 3 terms correct
(if using formula or completing the square allow one sign error and some simplification – allow as far as
\(\dfrac{7 \pm \sqrt{49 + 480}}{12}\) oe or
\(6\left(x - \dfrac{7}{12}\right)^2 - \dfrac{529}{24}\) oe or \(\left(x - \dfrac{7}{12}\right)^2 - \dfrac{529}{144}\) oe
A1: dep on M1 for two correct critical values
Accept –1.3……
May use \(\lt\), \(\leqslant\), \(\gt\) or \(\geqslant\) instead of =
M1ft: (dep on M1 and two critical values found)
for \(x \lt a\) and \(x \gt b\) where \(a\) is their lower critical value and \(b\) is their upper critical value
or \(x \gt \dfrac{5}{2}\) oe
or \(x \lt -\dfrac{4}{3}\) oe
or \(-\dfrac{4}{3} \gt x \gt \dfrac{5}{2}\) oe
A1: oe dep on previous M1
Accept –1.3…… or
\(\left(-\infty, -\dfrac{4}{3}\right), \left(\dfrac{5}{2}, (+)\infty\right)\) or \(\left(-\infty, -\dfrac{4}{3}\right) \cup \left(\dfrac{5}{2}, (+)\infty\right)\)
Do not ISW