Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$?

Question

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?

Answer 1

$x^2 - 2^2 = 0$

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

$x+2=0$ 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 $4$?

Answer 2

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

Answer 3

Just because

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

$$x=\sqrt4.$$

Answer 4

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

Answer 5

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 $\pm2$.

Answer 6

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.

Answer 7

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 $(\pm\sqrt.)$

Answer 8

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 $x - \sqrt{4} = 0 \iff x - 2 = 0$.

Answer 9

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!

Answer 10

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 $(-x)^2-4 = (-1)^2x^2-4 = x^4-4 = 0$

Answer 11

$$x=\sqrt{4}$$

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 $$x=2$$

Answer 12

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

Answer 13

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.