SOLUTION: The square of the first consecutive plus the square of the third consecutive integer equals 486 more than the square of the second consecutive integers. I TRY: x^2(x+2)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: The square of the first consecutive plus the square of the third consecutive integer equals 486 more than the square of the second consecutive integers. I TRY: x^2(x+2)      Log On


   

Question 468090: The square of the first consecutive plus the square of the third consecutive integer equals 486 more than the square of the second consecutive integers.



I TRY:
x^2(x+2)^2=(x+1)^2+486 (It was a mess when factor everything out)
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The square of the first consecutive plus the square of the third consecutive integer equals 486 more than the square of the second consecutive integers.
I TRY:
x^2(x+2)^2=(x+1)^2+486 (It was a mess when factor everything out)
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It says plus, you multiplied
x^2 + (x+2)^2=(x+1)^2+486
x%5E2+%2B+x%5E2+%2B+4x+%2B+4+=+x%5E2+%2B+2x+%2B+487
x%5E2+%2B+2x+-+483+=+0
(x - 21)*(x + 23 = 0
--> 21 & 23
or -23 & -21