SOLUTION: Find 3 consecutive odd integers such that twice the product of the first two is 7 more than the product of the last two. Find 2 consecutive integers whose product is 1 less tha

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Question 1086610: Find 3 consecutive odd integers such that twice the product of the first two is 7 more than the product of the last two.
Find 2 consecutive integers whose product is 1 less than their sum.
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
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Find 3 consecutive odd integers such that twice the product of the first two is 7 more than the product of the last two.
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Let the middle integer be n.
2%28n-2%29n=7%2Bn%28n%2B2%29
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2n%5E2-4n=n%5E2%2B2n%2B7
n%5E2-6n-7=0
%28n-7%29%28n%2B1%29=0
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n=-1, or n=7.

Do they both work?
2%28-1-2%29%2A%28-1%29=7%2B%28-1%29%28-1%2B2%29
-6%28-1%29=7-%281%29
6=6
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2%287-2%29%2A7=7%2B7%287%2B2%29
70=7%2B7%2A9
70=7%2B63
70=70
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Both solutions for n are good.

The three consecutive odd numbers:
-3, -1, 1
OR
5, 7, 9



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Find 2 consecutive integers whose product is 1 less than their sum.
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You try this one.