SOLUTION: Factorize 15000 in 2 consecutive numbers whose difference is 2

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Factorize 15000 in 2 consecutive numbers whose difference is 2       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 991564: Factorize 15000 in 2 consecutive numbers whose difference is 2
Found 2 solutions by Alan3354, Fombitz:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Factorize 15000 in 2 consecutive numbers whose difference is 2
----
Factoring implies integers.
121*123 = 14883
122*124 = 15128
No integer solution.
---
x*(x+2) = 15000
x%5E2+%2B+2x+-+15000+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-15000+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%282%29%5E2-4%2A1%2A-15000=60004.

Discriminant d=60004 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+60004+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+60004+%29%29%2F2%5C1+=+121.478569554024
x%5B2%5D+=+%28-%282%29-sqrt%28+60004+%29%29%2F2%5C1+=+-123.478569554024

Quadratic expression 1x%5E2%2B2x%2B-15000 can be factored:
1x%5E2%2B2x%2B-15000+=+%28x-121.478569554024%29%2A%28x--123.478569554024%29
Again, the answer is: 121.478569554024, -123.478569554024. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-15000+%29

===========

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Take the square root to get in the right neighborhood.
sqrt%2815000%29=122.5
15000=120%2A125
There are no consecutive factors of 15000.
120 and 125 are the factors closest to each other.