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Let's say we've got $\sqrt{4-3x} = x$.

Now let's say we have $x = 1,-4$.

Let's punch $-4$ in first.

We wind up with $\pm4 = -4$.

How is this an extraneous solution, but plugging in 1 isn't? Isn't a +- equal to negatives and positives?

David
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MrHost56
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    The most common convention, at least in calculus courses, is that $\sqrt{16}$ is the non-negative number whose square is $16$. Other conventions are possible. – André Nicolas Jun 18 '15 at 02:34

3 Answers3

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You need to be careful how you write the equation. I'm assuming it is $$\sqrt{4-3x}=x\ ,$$ as someone else has edited it.

In this case, remember that $\sqrt{a}$ denotes specifically the positive square root. It isn't plus-or-minus unless you actually write it as $\pm\sqrt a\,$. So substituting $x=-4$ gives $$\sqrt{16}=-4\ ,$$ that is, $$4=-4\ ,$$ which is not true.

David
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  • Don't you always do +- (forgive me for not using symbols, as I don't know how to do that on here) with square root? Because sqrt(16) can either be -4^2 or 4^2. – MrHost56 Jun 18 '15 at 02:27
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    With the square root symbol, as in $\sqrt{16}$, it is always positive. However that's not the same as solving $x^2=16$, in which case the answer is $x=\pm\sqrt{16}=\pm4$. – David Jun 18 '15 at 02:28
  • So only when you get a square root when solving for x the +- comes into play? – MrHost56 Jun 18 '15 at 02:29
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    I guess, though I'm not sure I would put it that way. The point is that some equations, particularly those involving powers, have more than one solution. – David Jun 18 '15 at 02:32
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Your statement $$\sqrt{4-3x} =x$$ can only be true when $x=1$

meiji163
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    That's what I don't understand. How is it only true if x = 1. I have four possible solutions ( x = -1,1,-4,4) that when plugged in all get x=x. – MrHost56 Jun 18 '15 at 02:25
  • @CollegeAnon The equation given has the principal root, without a $\pm$ – meiji163 Jun 18 '15 at 02:27
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Ok, here's the thing. A square root, when placed by itself, isn't negative, unless you put a "minus" in front of it. However, if your original equation was $x^2 = 4-3x$, both solutions would work. The confusion here arrives from you believing that $\sqrt{16}$ can be both positive and negative. Unless otherwise stated, $\sqrt{16}$ is always $4$, not $-4$, and so $4$ doesn't equal $-4$.

N. F. Taussig
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Kbot
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  • Why is that then? I don't understand how, logically, you can ever say it's positive because even if you're not solving for a variable, and it's just the square root by itself, it can still be negative or positive. Right? – MrHost56 Jun 18 '15 at 02:33
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    @CollegeAnon We do this so that $\sqrt{x}$ is a function. Functions only give us one output (here, the positive square root of $x$) for a given input (here, $x$). If $\sqrt{x}$ meant "either the positive, or negative, square root of $x$", then it wouldn't be a function. We do this because we like functions very much (and using $\pm$, we can still get the negative root if we need it). This answer might provide more context. – pjs36 Jun 18 '15 at 02:41
  • I kinda get what that guy is saying, though I don't understand how saying something is equal to both -3 and 3 means 3 = -3. Unless saying +-3 = -3 is like saying both 3 and -3 equal -3? – MrHost56 Jun 18 '15 at 02:55
  • Please see this tutorial on how to typeset mathematics on this site. – N. F. Taussig Jun 18 '15 at 08:42