SOLUTION: Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8. I know the answer is 3, 5, 7, but I need the work. Thanks!

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Question 338915: Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8. I know the answer is 3, 5, 7, but I need the work. Thanks!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
"Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8" means that x%28x%2B2%29=x%2B4%2B8


x%28x%2B2%29=x%2B4%2B8 Start with the given equation.


x%28x%2B2%29=x%2B12 Combine like terms.


x%5E2%2B2x=x%2B12 Distribute.


x%5E2%2B2x-x-12=0 Get every term to the left side.


x%5E2%2Bx-12=0 Combine like terms.


Notice that the quadratic x%5E2%2Bx-12 is in the form of Ax%5E2%2BBx%2BC where A=1, B=1, and C=-12


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%281%29+%2B-+sqrt%28+%281%29%5E2-4%281%29%28-12%29+%29%29%2F%282%281%29%29 Plug in A=1, B=1, and C=-12


x+=+%28-1+%2B-+sqrt%28+1-4%281%29%28-12%29+%29%29%2F%282%281%29%29 Square 1 to get 1.


x+=+%28-1+%2B-+sqrt%28+1--48+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-12%29 to get -48


x+=+%28-1+%2B-+sqrt%28+1%2B48+%29%29%2F%282%281%29%29 Rewrite sqrt%281--48%29 as sqrt%281%2B48%29


x+=+%28-1+%2B-+sqrt%28+49+%29%29%2F%282%281%29%29 Add 1 to 48 to get 49


x+=+%28-1+%2B-+sqrt%28+49+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


x+=+%28-1+%2B-+7%29%2F%282%29 Take the square root of 49 to get 7.


x+=+%28-1+%2B+7%29%2F%282%29 or x+=+%28-1+-+7%29%2F%282%29 Break up the expression.


x+=+%286%29%2F%282%29 or x+=++%28-8%29%2F%282%29 Combine like terms.


x+=+3 or x+=+-4 Simplify.


So the solutions are x+=+3 or x+=+-4


So the three consecutive odd integers are 3, 5, and 7


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website.

Jim