The sum of the squares of three consecutive positive numbers is 194. Find the
smallest number.
F4 Maths =.=
2009-10-15 5:24 am
回答 (2)
2009-10-15 5:28 am
✔ 最佳答案
Let the smallest number be x-1(x-1)^2 + x^2 + (x+1)^2 = 194
x^2 - 2x + 1 + x^2 + x^2 + 2x + 1 = 194
3x^2 + 2 = 194
x^2 = 192/3 = 64
x = 8 or - 8(rejected)
The smallest number is 8 - 1 = 7
2009-10-15 5:39 am
Let the smallest number is x.
x^2 + (x+1)^2 + (x+2)^2 = 194
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) =194
3x^2 + 6x + 5 =194
3x^2 + 6x -189 = 0
(3x - 21)(x + 9) = 0
x = 7 or -9
The smallest number is 7 or -9
x^2 + (x+1)^2 + (x+2)^2 = 194
x^2 + (x^2 + 2x + 1) + (x^2 + 4x + 4) =194
3x^2 + 6x + 5 =194
3x^2 + 6x -189 = 0
(3x - 21)(x + 9) = 0
x = 7 or -9
The smallest number is 7 or -9
參考: 自己
收錄日期: 2021-04-21 22:04:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091014000051KK01379