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$\frac{x^2+12x+36}{x^2+4x-12}=2$

I factored top and bottom to $\frac{(x+6)(x+6)}{(x+6)(x-2)}=2$ and eliminated the common factor (x+6) for $\frac{x+6}{x-2}=2$ then $x+6=2(x-2)$ and $x+6=2x-4$ and $6+4=2x-x$ and finally $x = 10$ Plugging this back into the original fraction proves true

My instructor however in her exam prep test says $\frac{x^2+12x+36}{x^2+4x-12}=2$

$$x^2+12x+36 = 2(x^2+4x-12)$$ $$x^2+12x+36 = 2x^2+8x-24$$ $$0 =2x^2-x^2 +8x -12x -24 -36$$ $$0 =x^2-4x -60$$ which she then factors to $(x-10)(x+6)$ saying the answer to the problem is $x=10$ or $x=-6$ But if you plug -6 back into the original fraction you get $0=2$. Where is the problem here? Thanks

user3255759
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    Simply put, your instructor has made a critical error, one that in my opinion calls into question her suitability as an instructor of mathematics. – heropup Mar 31 '17 at 01:43
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    The LHS is not defined for $x=-6$ and $x=2$. Hence, these do not count as roots of the given quantity. Technically speaking ,we search for roots in a certain "domain of definition". The domain of definition of the right hand side is precisely $\mathbb R $ without the points $-6$ and $2$, since at these points the function is undefined. Hence, since any equality between functions must respect domain, we have that the whole equality applies only when $x \neq -6,2$, so the rest of the argument must assume that $x \neq -6,2$ as well. This problem often crops up in trigonometric equations. – Sarvesh Ravichandran Iyer Mar 31 '17 at 01:44
  • I have over 40 pages of errors this professor of mathematics (university level precalc) has made on this course. I post to receive understanding of how she is incorrect so I might learn. I will have a chance to state my opinion of her suitability when I enter my course evaluation and submit my pages of error along with it. – user3255759 Mar 31 '17 at 01:49
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    If you plug -6 in, you get 0/0 =2 rather than 0=2. – Mark Joshi Mar 31 '17 at 02:50

3 Answers3

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You're right, for (almost) the very reason that you say: $-6$ is not a solution because the left-hand side is not defined when $x=-6$.

Matthew Leingang
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In short, you are correct.

Slightly longer, the expression $$\frac{x^2+12x+36}{x^2+4x-12}$$ is not defined for $x = {-6}$ because $x^2 + 4x -12 = 0$ when $x = -6$ and division by zero is undefined. Hence a priori to even talk about this fraction we assume that $x\neq -6$ and $x\neq2$.

Eff
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Other people have done an excellent job of pointing out why $-6$ cannot be a solution, but let me add exactly where your professor's approach breaks down. In the first step of cross-multiplying by $x^2+4x-12$, information is being "destroyed"; because you're multiplying by something that can be zero, you might be creating a $0=0$ situation and thus creating solutions that weren't present in the initial problem. Likewise, $\frac{2x}{x} = 1$ has no solution (because the left-hand side evaluates to $2$ whenever it is defined) but $2x = x$ has the new solution $x = 0$, corresponding to the manufactured $0=0$.

In fairness, though, this is an easy mistake to make, and one I wouldn't be surprised to see made by a perfectly competent teacher. If the numbers involved were slightly different, it would have turned out fine. I would argue that the real failure on the professor's part here was failing to check; a simple examination, as you noted, shows that $-6$ can't be a solution.

Reese Johnston
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  • Good answer (+1), and I agree that even a competent teacher could make such a mistake. However, apparantly OP has revealed in the comments that this is one of many mistakes, so it seems that she might not be completely competent. – Eff Mar 31 '17 at 04:02
  • Thank you for this further explanation. Sadly, in this prep test alone, even since I posted this request there have been three more errors including $6/-2=-2$ and $2x-8=5x+25$ becoming on the next line $2x+7x=25+8$ and $x^2+3x-10$ resulting in factors of $(x+5)(x-3)$ These errors I could see for myself without posting them on stack exchange. It has been a truly hair raising experience getting through this course. – user3255759 Mar 31 '17 at 04:40