Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 29, 2011"
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{{:AoPSWiki:Problem of the Day/June 29, 2011}} | {{:AoPSWiki:Problem of the Day/June 29, 2011}} | ||
==Solution== | ==Solution== | ||
− | + | First we have the question: <math>(ab+1)(a+1)(b+1)+ab</math> | |
+ | We multiply <math>(a+1)(b+1)</math> to get <math>(ab+1+a+b)</math> | ||
− | + | This makes the equation <math>(ab+1)(ab+1+a+b)+ab</math> | |
− | |||
− | |||
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− | This makes the equation <math>(ab+1)(ab+1+a+b)+ab<math> | ||
Now we seperate the equation to <math>(ab+1)(ab+1)+(ab+1)(a+b)+ab</math> | Now we seperate the equation to <math>(ab+1)(ab+1)+(ab+1)(a+b)+ab</math> | ||
− | We get <math>(ab+1)^2 +(a+b)(ab+1)+ab<math> | + | We get <math>(ab+1)^2 +(a+b)(ab+1)+ab</math> |
Now this is just a quadratic equation | Now this is just a quadratic equation | ||
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Thus we get a factored form of: | Thus we get a factored form of: | ||
− | <math>(ab+1+a)(ab+1+b)</math> | + | <math>\boxed{(ab+1+a)(ab+1+b)}</math> |
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Latest revision as of 17:48, 1 July 2011
Problem
AoPSWiki:Problem of the Day/June 29, 2011
Solution
First we have the question:
We multiply to get
This makes the equation
Now we seperate the equation to
We get
Now this is just a quadratic equation
Thus we get a factored form of: