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What is wrong with this solution?

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What is wrong with this solution? [#permalink]

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New post 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
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Re: What is wrong with this solution? [#permalink]

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New post 05 Mar 2012, 00:25
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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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Re: What is wrong with this solution? [#permalink]

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New post 05 Mar 2012, 04:02
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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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Kudos [?]: 5 [0], given: 6

Re: What is wrong with this solution? [#permalink]

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New post 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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