Question 353429: The sumof the squares of three consecutive positive integers is 77. Find the integers.
The book gives a hint: if one integer is x, the next consecutive positive integer is x+1, and the third is x+2
Found 3 solutions by stanbon, robertb, ikleyn:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The sum of the squares of three consecutive positive integers is 77. Find the integers.
The book gives a hint: if one integer is x, the next consecutive positive integer is x+1, and the third is x+2
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The "book" is correct, but you could just as well use, x-1,x,x+1
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Equation:
(x-1)^2 + x^2 + (x+1)^2 = 77
x^2 -2x + 1 + x^2 + x^2+2x+1 = 77
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3x^2 +2 = 77
3x^2 = 75
x^2 = 25
x = 5
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1st: x-1 = 4
2nd: x = 5
3rd: x+1 = 6
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Checking:
4^2 + 5^2 + 6^2
= 16 + 25 + 36
= 77
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Cheers,
Stan H.
Answer by robertb(5830)  (Show Source):
Answer by ikleyn(53619)  (Show Source):
You can put this solution on YOUR website! .
The sum of the squares of three consecutive positive integers is 77. Find the integers.
The book gives a hint: if one integer is x, the next consecutive positive integer is x+1, and the third is x+2
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There is much more elegant way to solve, than the way assumed in the book.
Let 'n' be the central number of these three consecutive integers.
So, our numbers are (n-1), n and (n+1).
Make an equation for the sum of squares
(n-1)^1 + n^2 + (n+1)^2 = 77,
(n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1) = 77,
Cancel opposite terms '-2n' and '2n' and combine like terms
3n^2 + 2 = 77 ---> 3n^2 = 77 - 2 ---> 3n^2 = 75 ---> n^2 = 75/3 = 25 ---> n =  = +/- 5.
Since we are looking for positive integer numbers, they are 4, 5 and 6. ANSWER
Solved, without necessity to solve a quadratic equation.
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