# Square root

1. Mar 6, 2010

### Taturana

Consider square root of 4.

Can square root of 4 be +2 or -2?

I asked that to my math teacher so he said: NO, square root of 4 is +2!

But I can't really understand why it cannot be -2 since -2 squared is also 4. One thing I've imagined is: square root of 4 can be +2 or -2, but for CONVENTION we use +2. Is that right?

There's another question related to that: consider

$$i^2 = -1$$

squaring the two sides we obtain

$$i^4 = 1$$

so

$$i = 1$$

so: can the i (the imaginary number) be 1? I think I'm making a lot of confusion with that, could someone explain me that?

Thank you

2. Mar 6, 2010

### eg2333

The square root of 4 can be +2 or -2, your math teacher is retarded. What you did with the imaginary number results in what I believe is called an extraneous solution (correct me if I'm wrong). If you substitute the 1 back into your original problem i^2=-1, you will find that 1=-1, which is not a true statement, meaning the solution is incorrect.

3. Mar 6, 2010

### jgens

We get these types of questions a lot so it might be worthwhile to run a forum search on this topic to get a very thorough review of it. In fact, I might even say that this question has been discussed ad nauseum

However, for your first question the, square root of a number is often defined as the principal (or positive) root. This convention is not without reason though; if we would like to have a square root function, the square roots must necessarily be single-valued. Therefore, $\sqrt{4} = 2$. For a more complicated example, let $x$ be a number such that $x^2 = 4$, then $\sqrt{x^2} = |x| = 2$ because $\sqrt{x^2}$ is necessarily positive. Since $|x| = 2$, this means that $x = 2$ or $x = -2$.

For your second question, squaring equations has the unfortunate effect of adding extraneous solutions. Taking your example, clearly $i^4 - 1 = 0$ in which case $(i^2 + 1)(i^2 - 1) = 0$. By the zero product property, either $i^2 + 1 = 0$ or $i^2 - 1 = 0$ (note that only one of these need hold). From the conventional definition of $i$ we have that $i^2 + 1 = 0$ and $i^2 - 1 = 0$ is just an extraneous solution added when we squared the equation.

Hopefully this all makes sense.

4. Mar 6, 2010

### Taturana

Thank you for the help guys. Now things make more sense.

PS: next time i'll do a forum search...

5. Mar 7, 2010

### HallsofIvy

If his teacher is retarded, what does that make you? His teacher is completely right. The equation $x^2= 4$ has two roots. But only one of them is $\sqrt{4}$ because $\sqrt{a}$ is defined as "the positive root of the equation $x^2= a$".

The reason we have to write the solutions to that equation as "$x= \pm\sqrt{a}$ is that $\sqrt{a}$ alone does NOT mean both solutions. If it did we would not need the "$\pm$".