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(x^2+6x+9)/(x+3) = 7

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AnnLuck
 Post subject: (x^2+6x+9)/(x+3) = 7
 Post Posted: Sat Jan 31, 2009 12:28 am 
2007 edition of Equations, Inequalities, & Vics
Chapter 3 Problem Set
Page 47, Question # 4

Give that (x^2+6x+9)/(x+3) = 7, what is x?

The solution in the study guide is to multiply each side by (x+3), set the equation to 0 and solve the quadratic equation. You then get that x=-3 or x=4. You then look at the problem and realize that x cannot equal -3 because the denominator would be 0.

My question is -- Why can't you factor the numerator, cancel out the (x+3) and then solve? Is it just by chance that it works for this equation? I know that the x squared means there are usually 2 answers (positive and negative), but if the denominator cancels out a factor in the numerator will the equation always have only one answer?
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JonathanSchneider
 Post subject: Re: (x^2+6x+9)/(x+3) = 7
 Post Posted: Thu Feb 19, 2009 2:38 pm 
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ManhattanGMAT Staff


Posts: 480
Location: Durham, NC
Actually, you can do that. And I agree with you that it's faster.
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