SOLUTION: find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.      Log On


   



Question 886344: find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
Put that into symbols.

n is any whole number.
The numbers needed are 2n+1, 2n+3, 2n+5.

The problem you are describing:
2%282n%2B1%29%282n%2B3%29=%282n%2B3%29%282n%2B5%29%2B7

Solve for n, and compute or evaluate each of the numbers.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.

Let first integer be F
Then 2nd is: F + 2, and 3rd is: F + 4
Therefore, 2(F)(F + 2) = (F + 2)(F + 4) + 7
2F%5E2+%2B+4F+=+F%5E2+%2B+6F+%2B+8+%2B+7
2F%5E2+-+F%5E2+%2B+4F+-+6F+-+15+=+0
Solve for F, the 1st integer, and then determine the other 2 integers. Note that 2 values will ensue for
the 1st integer, F, thus 2 values for the 2nd and 3rd integers.