
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 18,155 Thanks: 1422 
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: 1,659 Thanks: 652 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Quote:
It's almost done. Can you finish? Dan  
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: 2,824 Thanks: 752 
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.


Tags 
answer, correct, procedure, wrong 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Can someone confirm that this is correct or wrong...  perfect_world  Calculus  8  April 16th, 2015 05:13 AM 
Is my answer correct  KingsOfLeon  Probability and Statistics  2  December 6th, 2014 12:59 PM 
Is my answer correct ?  rsoy  Calculus  2  March 15th, 2013 05:55 PM 
Got this wrong on test . . .I think I'm correct  randomdude03  Linear Algebra  5  June 12th, 2012 01:13 PM 
i need urgent help with procedure as well as answer. please  adii shamz  Calculus  1  February 25th, 2009 03:18 PM 