Moderate -0.8 This is a straightforward surd comparison requiring a standard technique (squaring both sides) with minimal steps. The method is commonly taught and practiced, making it easier than average, though it does require knowing to square rather than attempting decimal approximation.
AG. OR e.g. \(\sqrt{3} \times \sqrt{3} \times \sqrt{2} > \sqrt{2} \times \sqrt{2} \times \sqrt{3}\), so \(\sqrt{3} > \sqrt{2}\), which is true. Allow proof that starts with answer and shows it must be true.
[3]
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| LHS is $(\sqrt{9} \times \sqrt{2}) = \sqrt{18}$ | B1 (2.1) | OR LHS squared is 18. No calculator. No decimal values allowed. |
| RHS is $(\sqrt{4} \times \sqrt{3}) = \sqrt{12}$ | B1 (1.1) | RHS squared is 12 |
| $\sqrt{18} > \sqrt{12}$ oe (so $3\sqrt{2} > 2\sqrt{3}$) | E1 (2.4) | AG. OR e.g. $\sqrt{3} \times \sqrt{3} \times \sqrt{2} > \sqrt{2} \times \sqrt{2} \times \sqrt{3}$, so $\sqrt{3} > \sqrt{2}$, which is true. Allow proof that starts with answer and shows it must be true. |
| **[3]** | | |