SOLUTION: FIND THREE CONSECUTIVE POSITIVE EVEN WHOLE NUMBERS SUCH THAT THE SUM OF THE SQUARE OF THE TWO SMALLER NUMBERS IS 20 MORE THAN THE SQUARE OF THE LARGEST NUMBER.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: FIND THREE CONSECUTIVE POSITIVE EVEN WHOLE NUMBERS SUCH THAT THE SUM OF THE SQUARE OF THE TWO SMALLER NUMBERS IS 20 MORE THAN THE SQUARE OF THE LARGEST NUMBER.      Log On


   

Question 250765: FIND THREE CONSECUTIVE POSITIVE EVEN WHOLE NUMBERS SUCH THAT THE SUM OF THE SQUARE OF THE TWO SMALLER NUMBERS IS 20 MORE THAN THE SQUARE OF THE LARGEST NUMBER.
Found 3 solutions by richwmiller, dabanfield, stanbon:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Stop shouting!
The numbers are n,n+2,n+4
n^2+(n+2)^2=(n+4)^2+20
We only want the positive solution.

Answer by dabanfield(803) About Me  (Show Source):
You can put this solution on YOUR website!
FIND THREE CONSECUTIVE POSITIVE EVEN WHOLE NUMBERS SUCH THAT THE SUM OF THE SQUARE OF THE TWO SMALLER NUMBERS IS 20 MORE THAN THE SQUARE OF THE LARGEST NUMBER.
Let x be the smallest of the three consecutive postive even integers.
Then the other two integers are x+2 and x+4.
We have then:
x^2 + (x+2)^2 = (x+4)^2 + 20
Solve for x and then calculate x+2 and x+4.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
FIND THREE CONSECUTIVE POSITIVE EVEN WHOLE NUMBERS SUCH THAT THE SUM OF THE SQUARE OF THE TWO SMALLER NUMBERS IS 20 MORE THAN THE SQUARE OF THE LARGEST NUMBER.
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1st: 2x-2
2nd: 2x
3rd: 2x+2
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Equation:
(2x-2)^2 + (2x)^2 = (2x+2)^2 + 20
4x^2 - 8x + 4 + 4x^2 = 4x^2 + 8x + 4 + 20
4x^2 - 16x -20 = 0
x^2 - 4x - 5 = 0
(x-5)(x+1) = 0
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Positive solution:
x = 5
1st: 2x-2 = 8
2nd: 10
3rd: 12
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Cheers,
Stan H.