The difference of two positive integers is 5.Two times the square of the smaller one exceeds the square of the larger one by71.Find the two integers.
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2012-09-20 2:35 am
回答 (2)
2012-09-20 2:52 am
✔ 最佳答案
Let a be the larger integer, and b be the smaller integer.a - b = 5 ..... [1]
2b² - a² = 71 ...... [2]
From [1] :
a = b + 5 ...... [3]
Put [3] into [2] :
2b² - (b + 5)² = 71
2b² - b² - 10b - 25 = 71
b² - 10b - 96 = 71
(b - 16)(b + 6) = 0
b = 16 or b = -6 (rejected)
Put b = 16 into [3] :
a = 16 + 5
a = 21
Hence, the two integers are 21 and 16.
參考: micatkie
2012-09-20 4:25 am
Let x be the smaller interger,
then x+5 be the larger interger.
2x^2 - (x+5)^2 = 71
2x^2 - (x^2+10x+25) = 71
2x^2 - x^2 - 10x - 25 - 71=0
x^2 -10x -96 = 0
(x-16)(x+6) = 0 (Using quatratic formula or cross-method)
x-16 = 0 or x+6 = 0
x = 16 or x = -6 (rejected)
∴ The smaller interger is 16
The larger integer
= 16 + 5
= 21
∴ The two integers are 16 and 21.
then x+5 be the larger interger.
2x^2 - (x+5)^2 = 71
2x^2 - (x^2+10x+25) = 71
2x^2 - x^2 - 10x - 25 - 71=0
x^2 -10x -96 = 0
(x-16)(x+6) = 0 (Using quatratic formula or cross-method)
x-16 = 0 or x+6 = 0
x = 16 or x = -6 (rejected)
∴ The smaller interger is 16
The larger integer
= 16 + 5
= 21
∴ The two integers are 16 and 21.
參考: 我都讀緊F.4,岩岩教完呢課
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