The sum of the squares of three consecutive positive integers is 245. What are the integers?

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- May 9th 2008, 01:12 PMblame_canada100sum of intergers
The sum of the squares of three consecutive positive integers is 245. What are the integers?

- May 9th 2008, 01:17 PMMoo
- May 9th 2008, 01:17 PMNeedHelp18
x

X + 1

x + 2

3x + 3 = 245

ps

I second wat Moo said - May 9th 2008, 02:05 PMSoroban
Hello, blame_canada100!

Quote:

The sum of the**squares**of three consecutive positive integers is 245.

What are the integers?

The three integers are: .$\displaystyle x,\:x+1,\:x+2$

The sum of their squares is: .$\displaystyle x^2 + (x+1)^2 + (x+2)^2 \:=\:245$

. . which simplifies to: .$\displaystyle 3x^2 + 6x - 240 \:=\:0 \quad\Rightarrow\quad x^2 + 2x - 80\:=\:0$

. . which factors: .$\displaystyle (x - 8)(x+10) \:=\:0$

. . and has the positive root: .$\displaystyle x \,=\,8$

Therefore, the integers are: .$\displaystyle 8,\:9,\:10$

- May 9th 2008, 02:07 PMo_O
Since we're dealing with integers, the solution x = -10 works as well.

x = -10; x + 1 = -9; x + 2 = - 8

Their squares should sum of to 245 as well - May 9th 2008, 02:17 PMNeedHelp18
Ohh thats right! I didnt read the squares part.

O.O... the OP said they are positive integers.