A S3 Maths Question

2009-08-10 11:17 pm
Consider the expression (√a+√b)^2 , where a and b are not perfect squares . Find two possible sets of a and b such that the value of expression is a rational number .
更新1:

If the value of the expression is a rational number and a , b are not perfect square integers, then ab must be a perfect square integer. ^呢句唔明,可唔可以解釋、證明? thx

更新2:

同埋我想問你點解要乘個p?

回答 (1)

2009-08-11 1:18 am
✔ 最佳答案
(√a+√b)^2
= a + b + 2√(ab)
If the value of the expression is a rational number and a , b are not perfect square integers, then ab must be a perfect square integer.
Let :
b = a * p^2 (where p is any integer)
then ab = a ( a * p^2) = (ap)^2 is a perfect square integer.
For example , when a = 2009 , p = 3
then b = 2009 * 3^2 = 18081
(√a+√b)^2 = (√2009 + √18081)^2 = 32144
when a = 5 , p = 1
then b = 5 * 1^2 = 5
(√a+√b)^2 = (√5 + √5)^2 = 20




2009-08-11 00:45:56 補充:
題目要求 a 和 b 都要是開方不盡的數(如果吾係駛鬼你做咩,求其代D開得盡的 a , b ,個expression就實係有理數啦,邊有咁易?)

唯有找些開不盡的 a 和 b , 如果 a = 3 , b = 5
個expression = 3 + 5 + √15 , 由於 3*5=15 開方不盡 ,
所以個expression就唔會係有理數(rational number )

2009-08-11 00:46:22 補充:
所以如果要個expression有理,關鍵條件就是 √ab 要開得盡,所以
『ab must be a perfect square integer.』不過想補充一點, a 和 b唔一
定要 integer,總之開方得盡就得,因為題目只要求有理數,如果要求整數才要 integer。

2009-08-11 01:01:38 補充:
點解要乘個 P^2 :

求其代 a , b不會保證 √ab 開得盡,所以要設 b = ap^2,(p是任何有理數)

則 √ab =√[ a*(ap^2)] = √[(a*p) * (a*p)] = a*p (開得盡)

我幾乎以另一個形式重答了你的問題,希望你看得明。


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