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 Dynamics November 4th, 2017 08:41 AM

Here is a problem:

Solve: (x^2 - 4) = x^2 - 10x + 16

Procedure 1:

factor both sides for (x+2)(x-2) = (x-8 )(x-2)
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?

 skipjack November 4th, 2017 08:47 AM

In the first method, dividing by x-2 isn't possible if x = 2 (because division by zero isn't defined). Hence, you should consider whether x = 2 is a solution.

 topsquark November 4th, 2017 08:54 AM

Quote:
 Originally Posted by Dynamics (Post 583560) Here is a problem: Solve: (x^2 - 4) = x^2 - 10x + 16 Procedure 1: factor both sides for (x+2)(x-2) = (x-8 )(x-2) divide out common facotrs (x+2) = (x -8 ) simplify 0 = -10 which is obviously wrong
If x = 2 then you get 0 = -10. So this procedure is only valid if \$\displaystyle x \neq 2\$. (The reason for this is that if x = 2 then you are dividing both sides by x - 2 = 2 - 2 = 0, which is, of course, impossible.)

Quote:
 Originally Posted by Dynamics (Post 583560) 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
As this procedure shows x = 2, which your first method shows is incorrect, you have a problem. (Always look for this sort of thing.)

It's almost done. Can you finish?

-Dan

 Dynamics November 4th, 2017 09:15 AM

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 (x-n) 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 (x-n) is valid for all x values except for x=n. Therefore, if we find that dividing by factor (x-n) 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 (x-n) to be invalid- namely, n. So, in a sense, dividing by 0 in the form of dividing by a factor (x-n) 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.

 Country Boy November 6th, 2017 09:39 AM

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.

 EvanJ November 22nd, 2017 05:36 PM

Quote:
 Originally Posted by Dynamics (Post 583564) Therefore, if we find that dividing by factor (x-n) 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 (x-n) to be invalid- namely, n.
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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