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November 4th, 2017, 09:41 AM  #1 
Newbie Joined: Nov 2017 From: NJ Posts: 2 Thanks: 0  Wrong Answer for Correct Procedure?
Here is a problem: Solve: (x^2  4) = x^2  10x + 16 Procedure 1: factor both sides for (x+2)(x2) = (x8 )(x2) divide out common factors (x+2) = (x 8 ) simplify 0 = 10 which is obviously wrong. Procedure 2: subtract all values on left from both sides for 0= x^2  10x + 16  x^2 + 4 simplify 0= 10x + 20 which means 10x=20 x=2 I don't understand why is the first procedure wrong while the second procedure is correct? Last edited by skipjack; November 4th, 2017 at 09:44 AM. 
November 4th, 2017, 09:47 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,554 Thanks: 1479 
In the first method, dividing by x2 isn't possible if x = 2 (because division by zero isn't defined). Hence, you should consider whether x = 2 is a solution.

November 4th, 2017, 09:54 AM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,663 Thanks: 653 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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November 4th, 2017, 10:15 AM  #4 
Newbie Joined: Nov 2017 From: NJ Posts: 2 Thanks: 0 
Hi! Thank you for your help! I think I understand your point. Because we cannot divide by zero, we know that dividing by some factor (xn) is only valid if x does not equal n, (and, conversely, that dividing by some factor (x+n) is only valid if x does not equal n). Therefore, we know that dividing by some factor (xn) is valid for all x values except for x=n. Therefore, if we find that dividing by factor (xn) renders an invalid result, or a contradiction, then x must be the only number which we understood to be that which would make the division of (xn) to be invalid namely, n. So, in a sense, dividing by 0 in the form of dividing by a factor (xn) can give you the correct answer (in a negative way) since it reveals the only number that could have led to such a contradictory result. 
November 6th, 2017, 10:39 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,942 Thanks: 797 
In general, you have to be careful anytime you divide both sides of an equation by something involving the "unknown", x, say. Once you have arrived at one or more values for x, check to be sure that value of x does not make the divisor 0.

November 22nd, 2017, 06:36 PM  #6 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 582 Thanks: 81  You say "only," but keep in mind that there can be many values of x that make the denominator = 0. For example, if the denominator was the product of (x+1)(x+2)...(x+100), every negative integer from 1 to 100 would make the denominator 0. The word "only" is true only if the degree of x is 1. Even if you don't have to know about fractions with x to the 2nd or higher power now, you will need it later.


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