-2

I am currently studying Algebra and I came across this equation. $\sqrt x = x- 2$. What I find intresting is that this equation has two roots. 1, 4. If you plug in 4. The equation is satisfied as L.H.S has $\sqrt 4 = \pm 2$ and R.H.S has $4 - 2 = 2$. Similarly, if you plug in 1 into the above equation you get the following, $\sqrt 1 = \pm 1$ and $1 - 2 = -1$. In both cases, the critera as L.H.S = R.H.S (as L.H.S can be either positive or negative and R.H.S is positive)

If I plot this in desmos. You can clearly see that on pluggin both $1, 4$, the equations (both L.H.S and R.H.S) intersects at 4 but do not intersect at 1. I can't find a reason to this as to why they dont intersect as $\pm 1 $ has $-1$ on L.H.S and R.H.S also has $1-2 = -1$. Can someone explain why?

Edit: I found a follow up after a little bit more looking. Here

Ayush
  • 105
  • 4
    For real numbers, by definition, $\sqrt{x}$ is the unique nonnegative number $r$ such that $r^2=x$. So $\sqrt{4}$ is equal to $2$, not to $\pm2$, and $\sqrt{1}$ is equal to $1$, not $\pm 1$. So $1$ is not a solution. Any solution must satisfy $x-2\geq 0$,or $x\geq 2$. – Arturo Magidin Sep 28 '22 at 02:23
  • 1
    Just to add to Arturo's (and bobeyt6's) point, whenever a positive or negative square root comes up, there is always a $\pm$ (or sometimes $\mp$) printed just outside the square root (e.g. in the quadratic formula). This is how we know to consider both roots. For the times where we only want one root or the other, we just use $\sqrt{\hspace{2mm}}$ and $-\sqrt{\hspace{2mm}}$. – Theo Bendit Sep 28 '22 at 02:41
  • It is not true $\sqrt{x}=x-2$ has roots $1$ and $4$. What is true is that $x=(x-2)^2$ has roots $1$ and $4$. Squaring an equation can introduce new roots. – anon Sep 28 '22 at 02:51
  • @ArturoMagidin Thank you! So say $\sqrt x = \pm y$. I understand $x$ needs to be non-negative i.e. $x >= 0$ otherwise $\sqrt x$ woluld be imaginary. However, $y$ on the other hand can be anything as long as it makes $x$ non-negative. This has two-cases. $(-1)^2=1$ and $(1)^2 = 1$. I seem to have trouble understanding to why $-1$ isn't valid even though squaring it results in a non-negative $x$ which still seems to be true to defination i.e. a a non-negative – Ayush Sep 28 '22 at 03:07
  • @runway44 Thank You! I didnt knew it. Does this happen with all operations or just this specific operation? – Ayush Sep 28 '22 at 03:11
  • No, it's not just $x$ that is nonnegative. The value of the square root is nonnegative. It is false that $\sqrt{x}=\pm y$ unless $x=y=0$ (in which case the plus/minus is superfluous). Again: $\sqrt{1}$ is NOT equal to $\pm 1$. It is equal to $1$ and only to $1$. It is false that "$y$ can be anything". NO. Don't thank me if you don't actually read what I write. – Arturo Magidin Sep 28 '22 at 03:28
  • 2
    Reversible operations do not change the set of solutions, but irreversible operations (ones in which it's not always possible to determine a unique input from a known output) tend to introduce new roots. – anon Sep 28 '22 at 03:37
  • 1
    @Ayush I said the answer had to be nonnegative. You did not reply asking why that was the case (as you did now). Instead you replied saying that you understood that the input had to be nonnegative, but that of course the answer didn't need to be, contradicting what I had written and thus demonstrating beyond a doubt that you did not in fact read it, which was rude of you. It is defined that way so it is a function, and so it satisfies $(\sqrt{x})^2 =x$, which is not necessarily true if you don't restrict the output. – Arturo Magidin Sep 28 '22 at 03:40
  • 4
    @Ayush You are confusing calculating the value of the (singular definite article) square root of $a$ with finding the solutions to the equation $x^2=a$. They are different things. – Arturo Magidin Sep 28 '22 at 03:43
  • @ArturoMagidin I apologize. Maybe I misinterpreted. English isn't my first language. So my original question was exactly as to "why is this the case with respect to the defination?" instead of the other. Could you please tell me why is $(\sqrt{-1})^2 =1$ false? – Ayush Sep 28 '22 at 03:57
  • @Ayush I already did. IF you allow the value of $\sqrt{1}$ to be either $1$ or $-1$, then use the value $1$ for the first $\sqrt{1}$ and the value $-1$ in the second to get $\sqrt{1}\sqrt{1}=1(-1)=-1\neq 1$. Or if you prefer, if $\sqrt{1}=\pm1$, then $(\pm1)(\pm1)$ requires four calclulations: $(1)(1)$, $(1)(-1)$, $(-1)(1)$, and $(-1)(-1)$. Only half of them equal $1$, the other half equal $-1$. Either way, you cannot guarantee that $\sqrt{1}\sqrt{1}$ equals $1$, which is a BIG problem. English is not my first language either (nor my second, nor my third). – Arturo Magidin Sep 28 '22 at 04:18
  • 1
    @Ayush So to make sure the square root is a function, and has the correct algebraic properties that yield an unambiguous value, we must pick only one of the two solutions to $x^2=a$ as THE value of $\sqrt{a}$. And we (humanity) have decided that we will always, uniformly, pick the nonnegative one. So the value of $\sqrt{a}$ (the output, not the input) will always be greater than or equal to zero. Period. – Arturo Magidin Sep 28 '22 at 04:25
  • 1
    The proposal of making the square root multi-valued also doesn’t really address the question either. We would have the left hand side of $\sqrt{x} = x-2$ would take two values, but the right hand side would take one. Is it really any more correct to "simplify” $\pm 1$ to $-1$ than it is to say $\sqrt{1}=1$? – Theo Bendit Sep 29 '22 at 04:22

1 Answers1

1

For all real numbers, $\sqrt{x}$ denotes the nonnegative number that squares to $x$. Otherwise, it would not be a function. Hence, $1$ will not work since $\sqrt1=1\ne 1-2=-1$.

Edit: This is different from the solutions to the equation $x^2=a$, in which the solutions are $x=\pm\sqrt a$. The square root function returns only nonnegative numbers since a function can only have one $y$-value for every $x$-value in its domain. Note that both the range and domain of square root are all nonnegative real numbers.

bobeyt6
  • 1,282
  • I have a question here on this part, "non-negative number that squares to $x$". If I say $r$ is the number that squares to $x$. Why is $r$ bounded to the non-negative range? I think it should be $x$ intead of $r$ as $-r * -r $ also equals a positive $x$ via basic $\pm$ rule. ? If the previous makes $x$ non-negative and not imaginary. Should'nt $-r$ also qualify here as one of the roots? I just can't understand why was it discarded. Also this shifts the defination you mentioned from 'the nonnegative number that squares to x' to 'the number that squares to nonnegative x'. – Ayush Sep 28 '22 at 03:24
  • 3
    @Ayush We require the value of the square root to be nonnegative because the square root is a function. Functions may not take more than one value on a single input. Don't confuse the unique value of $\sqrt{a}$ with the solutions of $x^2=a$. Those are related, but different, things. – Arturo Magidin Sep 28 '22 at 03:32
  • 2
    @Ayush do you want $(\sqrt{x})^2$ to equal $x$? I do, and if $\sqrt{1}$ can be either $1$ or $-1$, then that is not always true, as $\sqrt{1}\sqrt{1}$ could equal $(1)(-1)=-1\neq 1$. – Arturo Magidin Sep 28 '22 at 03:38