5
$\begingroup$

If the equation is $x-\sqrt 4=0$, then $x=2$.

If the equation is $x^2-4=0$, then $x=\pm 2$.

Why is it not $x=\pm 2$ in the first equation?

$\endgroup$
  • 8
    $\begingroup$ Now then... don't down-vote this question without leaving a comment. It looks reasonably good, and if you think it's not, then please spend a minute to explain it!!! $\endgroup$ – barak manos Dec 29 '16 at 9:46
  • 4
    $\begingroup$ Plug in $-2$ in the first question: $-2 - \sqrt 4 = 0 \implies -2 -2 = 0 \implies -4 = 0$, not true. $\endgroup$ – Ovi Dec 29 '16 at 15:18
  • 2
    $\begingroup$ 1st, 2nd, 3rd. I down-voted this question because this question is again on the well-trodden path of the positiveness of the square-root. That way to me it does not show any research effort on the OP's part. I think it is high time that a community-wiki question and answer is written for this specific topic: square root of a real is always non-negative. $\endgroup$ – Shraddheya Shendre Jan 15 '17 at 12:31
  • 1
    $\begingroup$ In fact the questions which I linked are directly concerned with this question, and there are others in the related list(like the "imposter" solution and false positive solutions of equations involving radicals) which again involve the same idea: square-root of a real is non-negative. $\endgroup$ – Shraddheya Shendre Jan 15 '17 at 12:36
  • 3
    $\begingroup$ @user401699. See my new answer $\endgroup$ – Harsh Kumar Jan 15 '17 at 12:41

13 Answers 13

4
$\begingroup$

When $a\ge 0$, the symbol $\sqrt[n] a$ is the non negative $n$th root of $a$.

$\endgroup$
13
$\begingroup$

$x^2 - 2^2 = 0$

$(x+2)(x-2)=0$

$x+2=0$ or $x-2=0$

$\sqrt4 = 2$

$x-\sqrt4=0$

$x=2$

Why can't the square root of $4$ be $-2$ instead of $2$, if $-2$ times $-2$ also equals $4$?

$\endgroup$
  • 3
    $\begingroup$ +1. Liked your answer the most amongst all. Hope it goes all the way up there. :) $\endgroup$ – Shraddheya Shendre Jan 15 '17 at 12:58
10
+100
$\begingroup$

I think you are confused in the square root function, actually the think making you confused is that what is the value of $\sqrt4$ which is $2$ but not $\pm2$, since the square root of any number is positive.

And the case in which the value of $\sqrt4$ is $\pm2$ is a quadratic equation, and the thing which happens here is $$x^2-4=0$$ $$x^2=4$$ $$x=\pm\sqrt4$$ hence, $$x=\pm2$$ but not $$x=\sqrt4\ne\pm2$$ And in case of your question:

The first equation can be written as $$x-2=0$$ $$\implies x=2$$ $$OR$$ $\color{red}{\text{Since the first equation is linear so it cannot have two roots.}}$

$\endgroup$
  • 3
    $\begingroup$ I think you take it to long, but thats good. $\endgroup$ – Lelouch vi Britannia Jan 1 '17 at 15:04
5
$\begingroup$

Just because

$$\sqrt4=2$$ and the equation says

$$x=\sqrt4.$$

$\endgroup$
  • 1
    $\begingroup$ This mean that the solution for the second equation is also $x=2$ (and only $x=2$). This is not what the OP has mentioned, so perhaps you should address that too. $\endgroup$ – barak manos Dec 29 '16 at 9:38
  • 2
    $\begingroup$ @AtulMishra: because $-2\ne\sqrt 4$. $\endgroup$ – Yves Daoust Dec 29 '16 at 9:40
  • 1
    $\begingroup$ Well, $x^2-4=0 \implies x^2=4 \implies x=\sqrt4$. If the last implication is wrong, and should be $x=\pm\sqrt4$, then it might be a good idea to point it out at this point. $\endgroup$ – barak manos Dec 29 '16 at 9:42
  • 1
    $\begingroup$ @AtulMishra: if $1^2=(-1)^2$, can you conclude that $1=-1$ ??? $\endgroup$ – Yves Daoust Dec 29 '16 at 9:43
  • 2
    $\begingroup$ @barakmanos: I would prefer the OP to understand that the first equation is a linear one, which has a single solution, and has no relation to the second. $\endgroup$ – Yves Daoust Dec 29 '16 at 9:44
4
$\begingroup$

The symbol $\sqrt{a}$ stands for the principal square root of $a$ which is always positive. Thus, the value of $x$ in the first equation is $2$.

The solution of the second equation, $x^2-4=0$, is given by \begin{align*} &\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[2ex] =\ &\frac{0\pm\sqrt{(0)^2-4(1)(-4)}}{2(1)} \end{align*} Simplyfing the above expression, we get two answers: $+\sqrt{4}$ and $-\sqrt{4}$.

That's why the solution of the second equation is $\pm2$.

$\endgroup$
4
$\begingroup$

Case 1-

$x^2 = 4$ is a quadratic equation and hence has two roots.

$x^2 = 4$

Then $x = \pm 2$

Case 2-

$x = \sqrt4$ is equation with degree one . Hence it can have only one root .

$x = \sqrt{4}$

Then $x = 2$


Also,

$\sqrt 4$ is always $2$, by definition. But if you know that the square of something is $4$, then that something is either $2$ or $-2$, and without more information it's impossible to tell which.

Explanation gave by Arthur in this post.

$\endgroup$
3
$\begingroup$

Actually while solving a quadratic equation we drop one step to short the answer or to save time whatever it may be.

THE STEP IS: $$x^2-4=0\tag{Step $1$}$$ $$x^2=4\tag{Step $2$}$$ $$x=\pm\sqrt4\tag{Step $3$}$$ $$x=\pm2\tag{Step $4$}$$ In our solution we drop the $3^{rd}$ step. Actually the thing is that as the $(+)$ sign change side as $(-)$ and $(\times)$ sign change side as $(\div)$. Similarly the $(^2)$[square] sign change side as $(\pm\sqrt.)$

$\endgroup$
  • 1
    $\begingroup$ Ok thats fine. +1. $\endgroup$ – user401699 Jan 15 '17 at 12:51
3
$\begingroup$

Obviously the answer to

$x - k = 0$ is $x = k$. One answer.

And easily, but not obviously, the answer to $x^2 - k^2 = 0$ is $x =\pm k$. Two answers (if we assume $k \ne 0$).

So Question number 1: Why does $x - k$ have one answer while $x^2 - m = 0$ has two (assuming $m > 0$)?

And obviously $x - \sqrt{4} = 0$ has solution $x = \sqrt{4}$ and $x^2 - 4=0$ has two solutions $x = \pm 2$.

So Question number 2: Why is $\sqrt{4} = 2$ and not $\sqrt{4} = \pm 2$.

The answer to Question number 1 is pretty clear: For any $x \ne 0$ we know $(-x)^2 = (-1)^2 x^2 = x^2 > 0$ and we also know that if $0 < a < b$ then $a^2 < b^2$ i.e. $a^2 \ne b^2$. And we also know that for and $m > 0$ that there must be some number $b>0$ so that $b^2 = m$. (Actually, that last one is not obvious at all. But we can take it as a given. It's a matter of knowing the reals are in a continuum but... we'll leave it for now.)

Those three bits of knowledge allow as to conclude: for any $m > 0$ there are two possible numbers so that $x^2 = m$ one is .... let's call it $b$ and the other is $-b$. So $x^2 - m = 0 \implies x^2 = m$ has two answers.

And $x - k =0 \implies x = k$ has only one.

So... we just have to answer Question number 2: Why is $\sqrt{4} = 2$ and not $\sqrt{4} = \pm 2$.

The answer to that is that the $\sqrt{}$ of a number $m$ is NOT defined to be "the number that when squared is $m$". If $m > 0$ then there are two such numbers and that definition to refer to a single number is ambiguous.

Instead the $\sqrt{}$ of a number $m$ is defined to be "the positive or zero number that when square is $m$". If $m \ge 0$ there is only one such number. If $m > 0$ then there are two such numbers that when squared are $m$ but only one of them is positive.

So if we have $x^2 - m = 0 \implies x^2 = m$ there are two answers. The positive one is $\sqrt{m}$. The negative one is $-\sqrt{m}$. Or in other words $x^2 - m = 0 \implies x = \pm \sqrt{m}$.

So the statement "$x - \sqrt{4} = 0$ is not "$x$ minus one or the other of the two numbers that when squared are $4$ is equal to zero". It is actually "$x$ minus one or the positive numbers that when squared is $4$ is equal to zero". Or in other words $x - \sqrt{4} = 0 \iff x - 2 = 0$.

$\endgroup$
3
$\begingroup$

Your confusion is understandable!

You are thinking: if the square root of some number $x$ is that number $y$ such that $y^2 = x$, then why is the solution to $x = \sqrt{4}$ different from $x^2 - 4 = 0$ ?!?

And the answer is: it wouldn't be different! That is, if the square root of $x$ would indeed be defined as "that number $y$ such that $y^2 = x$", then $\sqrt{4}$ could be either 2 or -2.

But, obviously, that is not how we look at $\sqrt{4}$ since we all know that that is just 2. This means: thinking of the square root of $x$ as "that number $y$ such that $y^2 = x$" is apparently not how we think about the square root.

OK, but why not? Well, notice that the whole "that number" part would be misleading in the first place: it suggest that there is one number with this property, but obviously there is not. So, if anything, we would have to say that the square root of $x$ is "any number such that $y^2 = x$"

... and we could have done so ...

... but we didn't. OK, but then we (you!) ask once again: why not?

Well, one simple reason is that we want the square root to act like a function, meaning that for any $x$, there is only one $y$ that is "the" square root of $x$. And we want functions, because functions are super useful: one thing in, and one thing out! Calculations can be done, etc. etc.

OK, but how can we ensure a function? Well, one thing we can do is to define the square root of $x$ as "that positive number $y$ such that $y^2 = x$"

... and that's exactly what we did...

and hence the difference between $x = \sqrt{4}$ and $x^2 - 4 = 0$

Of course, we could also have defined it as that negative $y$ such that $y^2 = x$" ... but in most practical cases, the positive one is the one you want, and the one that most often makes concrete sense in real life applications.

As a final reason for defining the square root of a number to be what it is, is that it allows us to explicitly distinguish between the two solutions to $x^2-2=0$, those being $\sqrt{2}$ and $-\sqrt{2}$: if the square root of 2 was any number that when squared gives you 2, then there no longer is a difference between $\sqrt{2}$ and $-\sqrt{2}$. In fact, it would not even be clear how many, and which specific, numbers would be solutions to $x^2-2=0$ if you said $x = \sqrt{2}$! Indeed, without the square root picking out a specific number, how would you refer to these different solutions? There probably is a way, but the square root function certainly makes our mathematical lives a lot easier for something like this!

$\endgroup$
2
$\begingroup$

The root function $f(x) = \sqrt{x}$ is defined to be nonnegative.

So $\sqrt{2} = 1.4...$, however the equation $x^2 - 4=0$ has to solutions:

If $x$ is such a solution, then so is $-x$ since $(-x)^2-4 = (-1)^2x^2-4 = x^4-4 = 0$

$\endgroup$
  • $\begingroup$ Might the downvoteres explain what is wrong? $\endgroup$ – flawr May 14 '17 at 15:14
2
$\begingroup$

$$x=\sqrt{4}$$

But $\sqrt{4}$ is positive because $\sqrt{x}$ is always positive. Visualizing the graph of $y=\sqrt{x}$ will reveal why in a few seconds.

Therefore $$x=2$$

$\endgroup$
2
$\begingroup$

The function $\sqrt. $ returns only non-negative numbers. And its range is [0,$\infty$) that's why $\sqrt4 = 2$ and $\sqrt4 ≠ ±2$

$\endgroup$
1
$\begingroup$

So the first problem basically reduces down to whether or not $\sqrt4=+2$ or if $\sqrt4=\pm2$, and my first initial response is this:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

And I ask "why would the $\sqrt\cdot$ have a $\pm$ in front if square roots were already taken to be $\pm$?"

And my answer would probably be "because $\sqrt\cdot$ is always positive by definition. It is not the case that one should even have $\sqrt\cdot$ be positive and negative simultaneously, or else:

  1. it wouldn't be a function (fails vertical line test)

  2. Things like substitutions in calculus would be very difficult if $\sqrt x$ was not a function

  3. And then when you trying to deal with complex numbers, everything explodes, since we are supposed to have $i=\sqrt{-1}$, but by your definition, $-i=\sqrt{-1}$ is also true, and then it all becomes so much more complicated.

Now, for the second problem, we can deduce that $x^2=4$, and since $(+x)^2=(-x)^2$, it follows that $x=\pm\sqrt4=\pm2$. Notice how this is contrary to square rooting. The differences are as follows:

  1. $x^2$ is a function, but it's inverse is not since it is not a bijective function.

  2. Positive and negative are taken since we have by definition $(-1)^2=1$, but it is also by definition that $-1\ne\sqrt1$, which many mistake to be true.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.