SOLUTION: find two consecutive odd integers whose product is 195

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: find two consecutive odd integers whose product is 195      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 258916: find two consecutive odd integers whose product is 195
Found 3 solutions by Greenfinch, stanbon, Alan3354:
Answer by Greenfinch(383) About Me  (Show Source):
You can put this solution on YOUR website!
Call the numbers 2n - 1 and therefore 2n+1
(2n-1)(2n+1)= 195
4n^2 - 196 = 0
2n = 14
So numbers are 13 and 15

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
find two consecutive odd integers whose product is 195
===
1st: 2x+1
2nd: 2x+3
-----
Equation:
(2x+1)(2x+3) = 195
4x^2 +8x + 3 = 195
4x^2 + 8x - 192 = 0
4(x^2 + 2x - 48)
(x-6)(x+8) = 0
---
Positive solution:
x = 6
1st: 2x+1 = 13
2nd: 2x+3 = 15
-------------------
Negative solution:
x = -8
1st: 2x+1 = -15
2nd: 2x+3 = -13
======================
Cheers,
Stan H.
======================

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Do these the easy way.
sqrt(195) is about 14.
One number is just below 14, the other above.
--> 13 & 15
-15 & -13 work also.