# Y=x^2, how do we know that x=-/+ sqrt(x) is always true?

1. Feb 7, 2009

### wajed

Y=x^2

I know that x= -/+ sqrt(y)

but I dont understand why I do this every time I try to solve "y=x^2" for "x".

I know the fact that -2^2=4, and also 2^2=4. and that is why we must have two answers, but I just need to understand how the process really goes, I just want to know the flaw of the logical steps that lead to the conclusion that x=-/+ sqrt(y).

The inverse of a function f(x) is defined as

I just dont get how hes done the last step!
huh, it really doesn't make sense to me, since he just applied the inverse -which in this case is sqrt. function- on the last step, which is the thing Im actually trying to understand how it happened!

PS: Im sorry, I havent studied mathematics thoroughly before, nor have I studied logic..
So, actually I don't know if I could put my question in a better way, because I dont even know where is the problem specifically,
Im just a high school student, trying to understand something out of his book, because he is burning to hell to know how this is happening.

2. Feb 7, 2009

### slider142

A function f from the real numbers into the real numbers can technically be interpreted as a list of ordered pairs of real numbers of the form (x, y) such that there is only one ordered pair in the set for each value of x. This is also interpreted graphically as the "vertical line test", where if you place a vertical line x=a on the graph of a function of x, it should only intersect the graph once. This definition is not abstract; in real world applications, many functions to be analyzed are lists of tabulated data from empirical observations, not neat equations (although they may be approximated by them).
An inverse function f-1 of a function f is a function that consists of a set where each ordered pair is reversed, only if such a set satisfies the criteria for being a function (passing the vertical line test). Graphically, you simply reflect the graph about the line y=x.
For the graph of the equation y = x2, we have a problem as the reflection of the graph (x = y^2) does not pass the vertical line test. In terms of ordered pairs, the original function y = f(x) = x2 is the list of ordered pairs of the form (x, x2) for all real numbers x. Reversing each ordered pair to form the inverse function f-1 gives us the list of all pairs of the form (x2, x). It is easy to see that this is not a function, as (1,1) and (-1, 1) get reversed to (1,1) and (1, -1). In other words, if some asked you, "what is f-1(1)?", you would not be able to give a unique answer, which is what functions are all about (each input should be associated with one output).
If we restrict our original function to the positive real numbers and 0, however, then reflecting the graph or reversing the ordered pairs does indeed yield an inverse function for f, which we denote by the principal square root function $f^{-1}(x) = \sqrt{x}$.
Now if y = x2 has the solution x = a, it is easy to show that x = -a is also a solution since (-1)(-1) = 1. Thus we have the normal solution $y = \pm\sqrt{x}$.

Last edited: Feb 7, 2009
3. Feb 7, 2009

### wajed

Im stupid that I had to read that text many times to get what you mean...

Ummm, may I ask you to show that x=-a is also a solution, please? (its not that Im lazy and dont want to think of it, Im just asking since it may make things clear)

maybe I didnt get it 100%, but anyway Ive added some information to consideration.

4. Feb 7, 2009

### HallsofIvy

Staff Emeritus
The product of two negative numbers is positive. In particular (-1)(-1)= 1. (-a)2= (-a)(-a)= (-1)(-1)(a)(a)= (1)(a2). I think that is what you are asking.

5. Feb 7, 2009

### Citan Uzuki

wajed, it's no wonder you're confused -- you've had two different people (the Y!A guy and slider142) give you a whole lot of background information without actually answering your question.

As HallsofIvy has mentioned, the ± in front of the square root symbol is necessary because while (by definition) (√y)² = y, it is also true that (-√y)² = (-1)²(√y)² = 1*y = y. So knowing only that x² = y, it might be that x=√y, but x² would also equal y if x=-√y. So there are at least two distinct possibilities. It turns out that these are the only two possibilities -- i.e. if x²=y, then either x=√y or x=-√y, which we abbreviate as x=±√y.

How do we know these are the only two possibilities? Well, consider the following: If x² = y, then x² - y = 0. Now, by definition, y = (√y)², so substituting we have that x² - (√y)² = 0. Factoring the left hand side, we have that (x - √y)(x + √y) = 0. The only way in which the product of two numbers can be zero is if at least one of the two factors is 0, so it must be the case that either x - √y = 0 or that x + √y = 0. In the first case, x=√y. In the second case, x=-√y. Thus we know x=±√y, and we are done.

6. Feb 7, 2009

### wajed

" The product of two negative numbers is positive. In particular (-1)(-1)= 1. (-a)2= (-a)(-a)= (-1)(-1)(a)(a)= (1)(a2). I think that is what you are asking. "

well, I understand that (-a)(-a)=a^2,

but anyway, that is my problem, that I don't know how to ask my questions..

now, actually I got it..
the answer is just a combination of two functions, 1st is restricted from [0,infinity) and second is [0,-infinity)

my question was just about proving that this is valid using logic (using pure mathematical terms)

so I think this is enough for now, or I`d just be asking for something that would need a big answer of a size of a book.

Thank you very much.

7. Feb 7, 2009

### HallsofIvy

Staff Emeritus
Right. More technically, because f(x)= x2 is not "one-to-one" (two different values of x give the same x2) it does not have a true inverse function. In order to get an "inverse" we have to restrict x2 to a domain in which it is one-to-one. As you say, we can do that by separating into two separate functions defined on $(-\infty,0]$ and $[0,\infty)$ since x= 0 is the "turning point". If f(x)= x2 for $0\le x< \infty$, then the inverse function is $f^{-1}(x)= \sqrt{x}$. If f(x)= x2 for $-\infty< x\le 0$, then the inverse function is $f^{-1}(x)= -\sqrt{x}$.

8. May 16, 2009

### wajed

Sorry for polling this up.
Just wanted to say that I read this again, and learned a alots.
Thank you.