Higher June 2025 Paper 2R Q21
21 Solve the inequality \(2x^2 - 7x - 15 \gt 0\)
Show clear algebraic working.
(3)
| Scheme | Marks |
|---|---|
eg \((2x + 3)(x - 5)\) oe or \(\dfrac{--7 \pm \sqrt{(-7)^2 - 4 \times 2 \times (-15)}}{2 \times 2}\) oe or \(2\left[\left(x - \dfrac{7}{4}\right)^2 - \left(\dfrac{7}{4}\right)^2\right] - 15\) oe | M1 |
| \((x =)\;-\dfrac{3}{2}\) and \((x =)\;5\) | A1 |
Working required Answer: \(x \lt -\dfrac{3}{2}\), \(x \gt 5\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for a correct method to find the critical values
Minimum evidence for quadratic formula is a two-term discriminant, eg \(\dfrac{7 \pm \sqrt{49 + 120}}{4}\) (must have the \(\pm\))
Allow \((x + 1.5)(2x - 10)\) as a correct factorisation, but do not allow \(\left(x + \dfrac{3}{2}\right)(x - 5)\) unless preceeded by division of the quadratic by 2
A1: dep on M1
for correct critical values oe
A1: dep on M1
for correct inequalities (must be separate inequalities)
allow interval notation
eg \(\left(-\infty, -\dfrac{3}{2}\right) \cup (5, \infty)\) or \(\left(-\infty, -\dfrac{3}{2}\right), (5, \infty)\)
or \(\left]-\infty, -\dfrac{3}{2}\right[ \cup \left]5, \infty\right[\) or \(\left]-\infty, -\dfrac{3}{2}\right[, \left]5, \infty\right[\)
Acceptable notation: allow a comma, space, “or”, “and” or “\(\cup\)” to link the two regions
Do not allow as a single inequality \(-\dfrac{3}{2} \gt x \gt 5\)
Allow M1A1 for the correct critical values AND evidence of another algebraic method that has led to these:
eg
\(x(2x + 3) - 5(2x + 3)\) and the correct critical values
or
\(2x(x - 5) + 3(x - 5)\) and the correct critical values
or
\((2x + 3)(2x - 10)\) and the correct critical values