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# What is wrong with this solution?

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Intern
Joined: 09 Oct 2011
Posts: 10
Location: United States
GMAT Date: 12-27-2011
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 5 [0], given: 6

What is wrong with this solution? [#permalink]  04 Mar 2012, 20:49
I used two different methods to solve an equation and I'm getting different answers. I'm not sure where am I going wrong.

Method 1 - Solving for x
$$x^2-16=x-4$$
$$x^2-x-12=0$$
$$(x-4)(x+3)=0$$
$$x=4, -3$$

Method 2 - Solving for x
$$x^2-16=x-4$$
$$x^2-4^2=x-4$$
$$(x+4)(x-4)=x-4$$
$$(x+4)=\frac{(x-4)}{(x-4)}$$
$$(x+4)=1$$
$$x=-3$$

It looks like something is wrong with method 2. Also in method 2 x cannot be 4 else the denominator will be 0.
Can someone tell me what am I doing wrong in method 2.

Thank you
Math Expert
Joined: 02 Sep 2009
Posts: 27228
Followers: 4231

Kudos [?]: 41108 [1] , given: 5666

Re: What is wrong with this solution? [#permalink]  05 Mar 2012, 00:25
1
KUDOS
Expert's post
vinayanand wrote:
I used two different methods to solve an equation and I'm getting different answers. I'm not sure where am I going wrong.

Method 1 - Solving for x
$$x^2-16=x-4$$
$$x^2-x-12=0$$
$$(x-4)(x+3)=0$$
$$x=4, -3$$

Method 2 - Solving for x
$$x^2-16=x-4$$
$$x^2-4^2=x-4$$
$$(x+4)(x-4)=x-4$$
$$(x+4)=\frac{(x-4)}{(x-4)}$$
$$(x+4)=1$$
$$x=-3$$

It looks like something is wrong with method 2. Also in method 2 x cannot be 4 else the denominator will be 0.
Can someone tell me what am I doing wrong in method 2.

Thank you

Second approach is wrong because what you are actually doing there is reducing (dividing) both parts of the equation by x-4, thus assuming with no ground for this that x-4 does not equal to zero (otherwise dividing by it wouldn't be allowed) hence you are excluding perfectly valid solution x-4=0 --> x=4.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Hope it's clear.
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Posts: 5460
Location: Pune, India
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Re: What is wrong with this solution? [#permalink]  05 Mar 2012, 04:02
1
KUDOS
Expert's post
vinayanand wrote:
I used two different methods to solve an equation and I'm getting different answers. I'm not sure where am I going wrong.

Method 1 - Solving for x
$$x^2-16=x-4$$
$$x^2-x-12=0$$
$$(x-4)(x+3)=0$$
$$x=4, -3$$

Method 2 - Solving for x
$$x^2-16=x-4$$
$$x^2-4^2=x-4$$
$$(x+4)(x-4)=x-4$$
$$(x+4)=\frac{(x-4)}{(x-4)}$$
$$(x+4)=1$$
$$x=-3$$

It looks like something is wrong with method 2. Also in method 2 x cannot be 4 else the denominator will be 0.
Can someone tell me what am I doing wrong in method 2.

Thank you

Check out this post for an explanation on when you can/cannot divide by a variable:
in-which-one-of-the-following-choices-must-p-be-greater-than-127621.html#p1048624
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Karishma
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Intern
Joined: 09 Oct 2011
Posts: 10
Location: United States
GMAT Date: 12-27-2011
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 5 [0], given: 6

Re: What is wrong with this solution? [#permalink]  05 Mar 2012, 04:58
Thank you Bunuel and Karishma
Re: What is wrong with this solution?   [#permalink] 05 Mar 2012, 04:58
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