Higher November 2022 Paper 1 Q20
20 The only solution to \(\quad x^2 + bx + c = 0 \quad\) is \(\quad x = -15\)
Work out the values of \(b\) and \(c\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((x + 15)^2\) | M1 | |
| \(x^2 + 15x + 15x + 225\) or \(x^2 + 30x + 225\) or \(b = 30\) or \(c = 225\) | M1dep | |
| \(b = 30\) and \(c = 225\) | A1 | |
| Alternative method 2: simultaneous equations using \(x = -15\) and \(b^2 - 4ac = 0\) | ||
| \((-15)^2 - 15b + c = 0\) or \(b^2 - 4\ (\times 1) \times c = 0\) | M1 | oe do not allow missing brackets unless recovered |
| \(b^2 - 4\ (\times 1) \times (15b - 225) = 0\) or \(b^2 - 60b + 900 = 0\) or \((b - 30)^2 = 0\) or \(b = 30\) or \(c = 225\) | M1dep | oe method to eliminate one unknown eg \(\left(\dfrac{225 + c}{15}\right)^2 - 4c = 0\) |
| \(b = 30\) and \(c = 225\) | A1 | |
| Alternative method 3: using \(b^2 - 4ac = 0\) in the quadratic formula | ||
| \(-15 = \dfrac{-b}{2(\times 1)}\) | M1 | oe |
| \(b = 30\) | M1dep | |
| \(b = 30\) and \(c = 225\) | A1 | |
Additional guidance
30 and 225 may come from incorrect working
eg do not allow \(c = 225\) from \((x - 15)^2\)