Higher November 2021 Paper 4 Q20
20 Solve algebraically.
\[\begin{aligned} y &= x + 3 \\ (x - 3)^2 + y^2 &= 50 \end{aligned}\]You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(x\)=] −4 [\(y\)=] −1 [\(x\)=] 4 [\(y\)=] 7 with correct algebraic working | 5 | accept any correct method M1 for correct substitution e.g. \((x - 3)^2 + (x + 3)^2\) [= 50] M1 for expanding both brackets correctly e.g. \(x^2 - 3x - 3x + 9 + x^2 + 3x + 3x + 9\) [=50] M1 for simplifying their equation e.g. \(2x^2 = 32\) or \(2x^2 - 32\) [= 0] | “Correct algebraic working” requires evidence of at least M1M1 implied by \(2x^2 + 18\) [= 50] condoning one error or better to \(ax^2 = b\) or to \(ax^2 + bx + c\) [= 0] |
| A1FT for \(x = -4, 4\) If 0 scored SC2 for [\(x\)=] −4 [\(y\)=] −1, [\(x\)=] 4 [\(y\)=] 7 with no working or SC1 for both \(x\) values with no working or a correct pair of \(x\) and \(y\) values with no working | FT their quadratic equation See appendix for alternative methods | ||
Appendix: Question 20
M1 for correct substitution
e.g. \((y - 3 - 3)^2 + y^2\) [= 50]
M1 for expanding the bracket correctly
e.g. \(y^2 - 6y - 6y + 36 + y^2\) [=50]
M1 for simplifying their equation
e.g. \(2y^2 - 12y - 14\) [= 0] or better e.g. \(y^2 - 6y - 7\) [= 0]
A1FT for \(y = -1, 7\)