SOLUTION: The sum of the squares of three consecutive, positive integers is equal to the sum of the squares of the next two integers. Find the five integers. x + x +2 + x + 3 = x squa

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: The sum of the squares of three consecutive, positive integers is equal to the sum of the squares of the next two integers. Find the five integers. x + x +2 + x + 3 = x squa      Log On


   

Question 195949: The sum of the squares of three consecutive, positive integers is equal to the sum of the squares of the next two integers. Find the five integers.

x + x +2 + x + 3 = x squared + x squared.. is the best I can come up with???? help! Thanks.
Found 2 solutions by jim_thompson5910, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Consecutive integers follow the form: x, x+1, x+2, x+3, etc...


So...

"The sum of the squares of three consecutive, positive integers is equal to the sum of the squares of the next two integers." translates to x%5E2%2B%28x%2B1%29%5E2%2B%28x%2B2%29%5E2=%28x%2B3%29%5E2%2B%28x%2B4%29%5E2


x%5E2%2B%28x%2B1%29%5E2%2B%28x%2B2%29%5E2=%28x%2B3%29%5E2%2B%28x%2B4%29%5E2 Start with the given equation.


x%5E2%2Bx%5E2%2B2x%2B1%2Bx%5E2%2B4x%2B4=x%5E2%2B6x%2B9%2Bx%5E2%2B8x%2B16 FOIL


3x%5E2%2B6x%2B5=2x%5E2%2B14x%2B25 Combine like terms.


3x%5E2%2B6x%2B5-2x%5E2-14x-25=0 Get all terms to the left side.


x%5E2-8x-20=0 Combine like terms.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=1, b=-8, and c=-20


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%28-8%29+%2B-+sqrt%28+%28-8%29%5E2-4%281%29%28-20%29+%29%29%2F%282%281%29%29 Plug in a=1, b=-8, and c=-20


x+=+%288+%2B-+sqrt%28+%28-8%29%5E2-4%281%29%28-20%29+%29%29%2F%282%281%29%29 Negate -8 to get 8.


x+=+%288+%2B-+sqrt%28+64-4%281%29%28-20%29+%29%29%2F%282%281%29%29 Square -8 to get 64.


x+=+%288+%2B-+sqrt%28+64--80+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-20%29 to get -80


x+=+%288+%2B-+sqrt%28+64%2B80+%29%29%2F%282%281%29%29 Rewrite sqrt%2864--80%29 as sqrt%2864%2B80%29


x+=+%288+%2B-+sqrt%28+144+%29%29%2F%282%281%29%29 Add 64 to 80 to get 144


x+=+%288+%2B-+sqrt%28+144+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%288+%2B-+12%29%2F%282%29 Take the square root of 144 to get 12.


x+=+%288+%2B+12%29%2F%282%29 or x+=+%288+-+12%29%2F%282%29 Break up the expression.


x+=+%2820%29%2F%282%29 or x+=++%28-4%29%2F%282%29 Combine like terms.


x+=+10 or x+=+-2 Simplify.


So the answers are x+=+10 or x+=+-2

Once again, the problem mentions that the numbers are positive. So this means that the only solution is x+=+10


This means that the numbers are: 10, 11, 12, 13, and 14

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The sum of the squares of three consecutive, positive integers is equal to the sum of the squares of the next two integers. Find the five integers.
x + x +2 + x + 3 = x squared + x squared.. is the best I can come up with???? help! Thanks.
------------
It's 3 consecutive integers:
x^2
(x+1)^2
(x+2)^2
= 3x^2 6x + 5
That's equal to (x+3)^2 + (x+4)^2
3x^2+6x+5 = x^2+6x+9 + x^2+8x+16
3x^2+6x+5 = 2x^2 + 14x + 25
x^2 - 8x - 20 = 0
(x-10)*(x+2) = 0
x = 10 (-2 is not positive)
100 + 121 + 144 =? 169 + 196
365 = 365
x = 10 works