Easy -1.8 This is a very straightforward question requiring only basic understanding of logical equivalence and finding a simple counterexample (x = -5 satisfies x² = 25 but not x - 5 = 0). It requires minimal calculation and tests only the most basic logical reasoning, making it significantly easier than typical A-level questions.
Mention of \(-5\) as a square root of 25 or \((-5)^2 = 25\)
M1
Condone \(-5^2 = 25\)
\(-5 - 5 \neq 0\) o.e. or \(x + 5 = 0\)
M1
or, dep on first M1 being obtained, allow M1 for showing that 5 is the only solution of \(x - 5 = 0\). Allow M2 for \(x^2 - 25 = 0\), \((x+5)(x-5)[=0]\) so \(x - 5 = 0\) or \(x + 5 = 0\)
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mention of $-5$ as a square root of 25 or $(-5)^2 = 25$ | M1 | Condone $-5^2 = 25$ |
| $-5 - 5 \neq 0$ o.e. or $x + 5 = 0$ | M1 | or, dep on first M1 being obtained, allow M1 for showing that 5 is the only solution of $x - 5 = 0$. Allow M2 for $x^2 - 25 = 0$, $(x+5)(x-5)[=0]$ so $x - 5 = 0$ or $x + 5 = 0$ |
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