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
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
