Higher June 2025 Paper 2 Q15
15 Solve \(\quad 3x^2 + 5x - 9 = 0\)
Give your solutions as decimals. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{-5 \pm \sqrt{5^2 - 4 \times 3 \times -9}}{2 \times 3}\) or \(\pm\sqrt{\left(\dfrac{5}{6}\right)^2 + 3} - \dfrac{5}{6}\) | M1 | oe eg \(\dfrac{-5 \pm \sqrt{133}}{6}\) may be seen as two separate calculations |
| [1.08, 1.09] or 1.1 and [−2.76, −2.75] or −2.8 | A1 | two decimal solutions in either order |
Additional guidance
| One value correct from T & I | M0 |
| Only using one of \(+\) or \(-\) instead of \(\pm\) is M0 unless recovered | |
| \(\sqrt{5}^2 - 4 \times 3 \times -9\) is M0 unless recovered | |
| \(-\dfrac{5}{6} \pm \sqrt{5^2 - 4 \times 3 \times -9}\) is M0 unless recovered | |
| Both decimal solutions seen in working but only one on answer line | M1A0 |
| Both decimal solutions seen in working but signs transposed on answer line | M1A0 |