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This question was brought up by a high school math teacher as a question on the SAT! I think I found the error in his logic which I wrote below, but I just wanted to be sure.

$\underline{\text{Note:}}$ Equation $(6)$ is a false conclusion to make as $x=3$ is not valid for equation $(1)$.

Assume equation $(1)$ one holds.

$\sqrt{x-2}=x-4\tag{1}$

$\sqrt{x-2}^2=(x-4)^2\tag{2}$

$x-2=x^2-8x+16\tag{3}$

$0=x^2-9x+18\tag{4}$

$0=(x-6)(x-3)\tag{5}$

$\text{So, 6 and 3 should be valid solutions to equation (1)} \tag{6}$

I think the reason for this error is that equation $(6)$ should be $x=6$ or $x=3$ instead. Since substitution of $3$ into equation $(1)$ is false, we know $x=6$. I just wanted to make sure that my reasoning was correct though.

W. G.
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    A priori, there might be both $3$ and $6$ as solutions to (1). It's possible that both solutions work, and neither is extraneous. However, when we check by plugging in, we see that $3$ is not a solution, and is extraneous. $6$ works. – vadim123 Jan 26 '19 at 01:52
  • That's what I thought as well. – W. G. Jan 26 '19 at 01:54
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    $\sqrt{x -2} = x-4\implies x-2 = 4 \ge 0$. So $x-2 = (x-4)^2$ AND $x \ge 2$ and $x \ge 4$. so $0 = x^2 -9 x + 18$ AND $x \ge 2$ and $x \ge 4$. and so $(x-3)(x-6) =0$ AND $x \ge 2$ and $x\ge 4$. so $x = 3$ or $x = 9$ AND $x \ge 2$ and $x \ge 4$. .... So $x = 9$. The end. – fleablood Jan 26 '19 at 02:03
  • However if you were given either $\pm \sqrt{x-2} = x-4$ of $\sqrt{x-2} = |x-4|$ or $M= x-4$ and $M^2 = x-2$ then $6$ and $3$ will both be valid answers. – fleablood Jan 26 '19 at 02:06
  • I see what you are doing! Equation $(1)$ implies $x\geq 4$. That makes sense. Thank you for your help! – W. G. Jan 26 '19 at 02:10
  • Should be solutions does not mean the same thing as are solutions. It might have been clearer if he had said So 3 and 6 are the only possible solutions. – Steven Alexis Gregory Jan 26 '19 at 02:26

2 Answers2

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$\sqrt{x-2} \ge 0$ by definition.

So you have $\sqrt{x-2} = x-4 \ge 0$. So you have the information $x \ge 4$.

When you square both sides $x - 2 = (x-4)^2$ you lose that information so you must make a note to keep it.

So you $x-2 = (x-4)^2; x \ge 4$.

Keep going you end up with

$(x-6)(x-3) =0; x \ge 4$ so you final conclusion is:

$x =6$ or $x =3$ but $x \ge 4$.

So $x = 6$.

Extraneous solutions come in when you lose or ignore information you really already know. $\sqrt{x-2} = x-4$ is not the whole story. You also know $x -2 \ge 0$ and $x- 4 \ge 0$. Don't ignore that you know that. So a more accurate story is $\sqrt{x-2} = x-4 \ge 0$.

fleablood
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Your reasoning is pretty good. To see exactly where the provided solution goes wrong:

Although it does necessarily follow from $\sqrt{a}=b$ that $a = b^2$, it does not necessarily follow from $a=b^2$ that $b=\sqrt{a}$ (because $b=-\sqrt{a}$ is also a solution).

In the deliberately wrong solution that they provide (using $a=x-2$ and $b=x-4$):

$\sqrt{x-2}=x-4$ does imply that $x-2=(x-4)^2$, but $x-2=(x-4)^2$ does not imply that $\sqrt{x-2}=x-4$ (because another solution is $-\sqrt{x-2}=x-4$). Thus, by solving for $x-2=(x-4)^2$, we are also solving for $-\sqrt{x-2}=x-4$, which was not the given information.

In short we need to be careful when squaring both sides of an equation to solve it. In doing so, we potentially destroy some of the information we are given.

Richard Ambler
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