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
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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) (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
Start with the given equation.
FOIL
Combine like terms.
Get all terms to the left side.
Combine like terms.
Notice we have a quadratic equation in the form of where , , and
Let's use the quadratic formula to solve for x
Start with the quadratic formula
Plug in , , and
Negate to get .
Square to get .
Multiply to get
Rewrite as
Add to to get
Multiply and to get .
Take the square root of to get .
or Break up the expression.
or Combine like terms.
or Simplify.
So the answers are or
Once again, the problem mentions that the numbers are positive. So this means that the only solution is
This means that the numbers are: 10, 11, 12, 13, and 14
Answer by Alan3354(69443) (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
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