Standard +0.3 This is a straightforward algebraic manipulation requiring rationalizing the denominator by multiplying by the conjugate (1 - √(2n)), then simplifying. It's a standard technique with clear steps, slightly easier than average as it's purely procedural with no problem-solving insight needed, though the algebra requires care.
Question 7:
7 | 3 + 8 n
1 + 2 n
3 + 8 n 1 − 2 n
=
1 + 2 n 1 − 2 n
3 − 3 2 n + 8 n − 1 6 n 2
=
1 − 2 n
3 − 3 2 n + 2 2 n − 4 n
=
1 − 2 n
3 − 2 n − 4 n
=
1 − 2 n
4 n − 3 + 2 n
=
2 n − 1
Simplifies 8 n to 2 2 n | 1.1b | B1
1− 2n
Multiplies by or
1− 2n
2 n − 1
2 n − 1 | 1.1a | M1
Obtains correct single fraction
with denominator of 1 − 2 n or
2n−1 | 1.1b | A1
Completes reasoned argument
4 n − 3 + 2 n
to obtain
2 n − 1
AG
Condone
eg 8 n or √2n except if seen on
their final line. | 2.1 | R1
Question 7 Total | 4
Q | Marking instructions | AO | Marks | Typical solution
Show that
$$\frac{3 + \sqrt{8n}}{1 + \sqrt{2n}}$$
can be written as
$$\frac{4n - 3 + \sqrt{2n}}{2n - 1}$$
where $n$ is a positive integer.
[4 marks]
\hfill \mbox{\textit{AQA Paper 1 2024 Q7 [4]}}