SOLUTION: Find three consecutive numbers where the product of the smallest two numbers is 28 less than the square of the largest numbers.

Question 314850: Find three consecutive numbers where the product of the smallest two numbers is 28 less than the square of the largest numbers.
Found 2 solutions by unlockmath, nerdybill:
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
Let's have x be one number and x+1 be the next and x+2 be the third. Now we can set up the following equation:
x(x+1)=(x+2)^2-28 (Hint: the word "is" in word problems means "equal" usually)
x^2+x=x^2+4x+4-28 Subtract x^2 from both sides and subtract 4x to get:
-3x=-24 Divide -3 into both sides to get:
x=8
So our numbers are 8,9,10
Make sense?
RJ
www.math-unlock.com
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Find three consecutive numbers where the product of the smallest two numbers is 28 less than the square of the largest numbers.
.
Let x = smallest of three consecutive numbers
then
x+1 = middle number
x+2 = largest number
.
(x+1)(x+2) = x^2 - 28
x^2+2x+x+2 = x^2 - 28
x^2+3x+2 = x^2 - 28
3x+2 = -28
3x = -30
x = -3 (smallest number)
.
middle:
x+1 = -3+1 = -2 (middle)
.
Largest number:
x+2 = -3+2 = -1 (largest)
.
solution, -3, -2, -1
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