# Quod Erat Demonstrandum

## 2009/10/18

### 致 5E 同學

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 1:40 下午

## 3 則迴響 »

1. 我唔合格, 仲要低過mean…

迴響 由 FTL — 2009/10/19 @ 11:56 下午 | 回應

2. Here is my way. (I didn’t read your marking, this may coincide with what you did.)

Since x + 4y = 0 is the tangent to 3x + (y – 1)^3 = 4. We have 3(-4y) + (y – 1)^3 = 4 has a repeated root.
Hence we have: -12 + 3(y – 1)^2 = 0. Solving yields: y = 3 or -1.
However, y = 3 does not satisfy the original equation, therefore the repeated root is -1.
The rest is trivial.

迴響 由 koopakoo — 2009/10/20 @ 1:46 上午 | 回應

• Thank you Koopa, it is a trick appears in the topic of polynomials in pure mathematics syllabus, I did not used this method in the F.5 quiz as expected. Thank you again!

迴響 由 johnmayhk — 2009/10/20 @ 12:47 下午 | 回應