SOLUTION: Find 3 consecutive even numbers where the product of the smaller two numbers is 40 less than the square of the largest number.

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Question 282198: Find 3 consecutive even numbers where the product of the smaller two numbers is 40 less than the square of the largest number.
Found 2 solutions by Mathematicians, oberobic:
Answer by Mathematicians(84) About Me  (Show Source):
You can put this solution on YOUR website!
The key point is setting up the equation:
Let X be the smallest even integer solution
X + 2 will be the next even integer
X + 4 will be the third even integer
So the product of the two smaller ones is 40 less than the square of the largest number.
Here is how that equation forms out:
x+%2A+%28x+%2B+2%29+=+%28x+%2B+4%29%5E2+-+40
x%5E2+%2B+2x+=+%28x%5E2+%2B+8x+%2B+16%29+-+40 distribute and foil
-6x+=+-24 combining like terms
x+=+4
if x = highlight%284%29, then the next two even integers are highlight%286%29 and highlight%288%29

Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
Three consecutive numbers could be called: a, b, c.
But it is far better to define them as:
a+=+x
b+=+x%2B1
c+=+x%2B2
.
a%2Ab+=+c%5E2+-40 :: the product of the two smaller is 40 less than the square of the largest
.
x%28x%2B1%29+=+%28x%2B2%29%5E2+-40
x%5E2%2Bx+=+x%5E2%2B4x%2B4+-40
x%5E2%2Bx+-x%5E2+-4x+=+-36
-3x+=+-36
x+=+12
x+%2B+1+=+13
x+%2B+2+=14
.
Checking...
12%2A13+=+156
14%2A14+=+196
So
12%2A13+=+14%5E2+-+40
Correct.
.
Answer:
The three consecutive numbers are: 12, 13, and 14.
.
Done.