# MAT - Paper 2008 - Q2

> Source: https://ollybritton.com/notes/maths/problems/mat/2008/q2/ · Updated: 2021-10-13 · Tags: mat, notes, maths

## Flashcards
### 2021-10-13
##### You've rearranged an expression to get $$(8 - 2a^2)x^2 + (24 - 4ab)xy + (18-2b^2)y^2 = 0$$ What must now be true for it to always equal zero for positive integers $x$ and $y$??
$$
2a^2 = 8
$$
$$
4ab = 24
$$
$$
2b^2 = 18
$$

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