4400 Higher June 2006 Paper 4 Q17
17 Solve the equations
\(y = 2x + 1\)
\(x^2 + y^2 = 13\)
(6)
| Scheme | Marks |
|---|---|
| \(x^2 + (2x + 1)^2 = 13\) | M1 |
| \(x^2 + 4x^2 + 2x + 2x + 1 = 13\) \((5x^2 + 4x - 12 = 0)\) | M1 |
| \((5x - 6)(x + 2) = 0\) or \(x = \dfrac{-4 \pm \sqrt{4^2 - 4 \times 5 \times (-12)}}{2 \times 5}\) | M1 |
| \(x = -2\) and \(x = 1.2\) | A1 |
| Subst two values of \(x\) into eqn | M1 |
| \(x = -2\) & \(y = -3\) \(x = 1.2\) & \(y = 3.4\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: or further simplified
M1: condone without “= 0” oe \(\quad\) must be correct
M1: dep M2 For incorr \(x\) must see wking
A1: paired, eg by alignment or coords
T & I: 6 mks or 0 mks
Follow similar scheme for subst for \(x\)