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.

Question 256169: 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.
Found 2 solutions by solver91311, checkley77:
Answer by solver91311(24713) (Show Source): 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

Answer by checkley77(12844) (Show Source): 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

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