Higher June 2022 Paper 2 Q18
18 Solve \(\quad x^2 + 7x - 11 = 0\)
Give your solutions as decimals. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{-7 \pm \sqrt{7^2 - 4 \times 1 \times -11}}{2 \times 1}\) or \(-\dfrac{7}{2} \pm \sqrt{\left(\dfrac{7}{2}\right)^2 + 11}\) | M1 | oe eg \(\dfrac{-7 \pm \sqrt{49 + 44}}{2}\) or \(\dfrac{-7 \pm \sqrt{93}}{2}\) or \(-\dfrac{7}{2} \pm \sqrt{\dfrac{49}{4} + 11}\) or \(-\dfrac{7}{2} \pm \sqrt{\dfrac{93}{4}}\) |
| 1.3(2…) and \(-8.3(2\ldots)\) | A1 |
Additional guidance
| \(-3.5 \pm \sqrt{12.25 + 11}\) or \(-3.5 \pm \sqrt{23.25}\) | M1 |
| For M1 allow solutions given separately eg \(\dfrac{-7 + \sqrt{93}}{2}\) and \(\dfrac{-7 - \sqrt{93}}{2}\) | M1 |
| Both solutions correct | M1A1 |
| One solution correct does not imply M1 | |
| Not using \(\pm\) is M0 unless recovered | |
| eg1 \(\dfrac{-7 + \sqrt{7^2 - 4 \times 1 \times -11}}{2 \times 1}\) followed by 1.32 | M0A0 |
| eg2 \(\dfrac{-7 + \sqrt{7^2 - 4 \times 1 \times -11}}{2 \times 1}\) followed by 1.3 and \(-8.3\) | M1A1 |
| A short dividing line or a short square root symbol is M0 unless recovered eg by a correct solution | |
| Condone if their square root symbol is above any part of \(-11\) | |
| √(\(7^2 - 4 \times 1 \times -11\)) is correct for \(\sqrt{7^2 - 4 \times 1 \times -11}\) | |
| Both decimal solutions seen in working but only one on answer line | M1A0 |