Standard +0.3 This is a straightforward modulus equation requiring the standard technique of considering critical points and cases. While it involves a parameter 'a', the algebraic manipulation is routine: squaring both sides gives (x+a)² = (2x-5a)², leading to a simple factorization. This is slightly above average difficulty only because of the parameter, but remains a standard textbook exercise with no novel insight required.
State or imply non-modulus equation \((x+a)^2 = (2x-5a)^2\) or pair of linear equations
B1
SR B1 for \(x = 6a\)
Attempt solution of quadratic equation or of pair of linear equations
M1
Allow M1 if \(\frac{4}{3}\) and 6 seen
Obtain, as final answers, \(6a\) and \(\frac{4}{3}a\)
A1
Total:
3
## Question 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modulus equation $(x+a)^2 = (2x-5a)^2$ or pair of linear equations | B1 | SR **B1** for $x = 6a$ |
| Attempt solution of quadratic equation or of pair of linear equations | M1 | Allow **M1** if $\frac{4}{3}$ and 6 seen |
| Obtain, as final answers, $6a$ and $\frac{4}{3}a$ | A1 | |
| **Total:** | **3** | |
---