SOLUTION: Polynomial word problems help please! Find three consecutive odd integers if the product of the first and second integer is 7 less than 10 times the third integer.

You can put this solution on YOUR website!


Let represent the smallest of the three consecutive odd integers. The next larger consecutive odd integer must then be , and the one after that .

The product of the first and second must be: .

7 less than 10 times the third must be:

So:





Solve the easily factorable quadratic. Both roots are valid answers to the question.

John

You can put this solution on YOUR website!
LET X, X+2 & X+4 BE THE 3 INTEGERS.
X(X+2)=10(X+4)-7
X^2+2X=10X+40-7
X^2+2X-10X-33=0
X^2-8X-33=0
(X-11)(X+3)=0
X-11=0
X=11 ANS. FOR THE SMALLEST INTEGER.
11+2=13 ANS. FOR THE MIDDLE INTEGER.
11+4=15 FOR THE LARGEST INTEGER.
PROOF:
11*13=10*15-7
143=150-7
143=143