# Solution to a simple equation

1. Sep 1, 2012

1. The problem statement, all variables and given/known data

X^2 + 1 = 0

2. Relevant equations

find the solution

3. The attempt at a solution

As simple as this:

what is the solution to x^2 +1 = 0

.... my question, why am I wrong in thinking x^2 = -1 as a possible solution?

2. Sep 1, 2012

### phinds

X^2 = -1 is not a solution for "what is x", which is what is being asked for, it is a solution to "what is X^2"

3. Sep 1, 2012

Oh I see.... Well, a solution to x^2 could be $$\pm 1$$. So what is the solution, I am a bit confused. Would it be

x = -1/2

4. Sep 1, 2012

### Staff: Mentor

(-1/2)² = 1/4, so that's not a solution to X² + 1 =0

You are right in re-arranging your equation as X² = -1
now take the square root of both sides.

5. Sep 1, 2012

### micromass

Staff Emeritus
Some information is missing here. What is x supposed to be?? Is it rational? Real? Complex?
Whether or not a solution exists to this problem depends on what you allow x to be.

6. Sep 2, 2012

### HallsofIvy

Staff Emeritus
No, it's not. You have already said that x^2= -1, not $\pm 1$. Now you need to get x itself, not x^2. You need to "undo" the square- what's the opposite of squaring?

You seem to be confused about what "solution to an equation" means. What level mathematics are you taking? Where did you get this problem? Have you studied complex numbers.

7. Sep 3, 2012

So the solution is

$$x = \sqrt{-1}$$

8. Sep 3, 2012

Bravo.

9. Sep 3, 2012

### Mentallic

$$x=\pm\sqrt{-1}$$

10. Sep 3, 2012

### eumyang

To the OP: you know the symbol we use for $\sqrt{-1}$, right?

11. Sep 3, 2012

Yes, it's the imaginary number. :)

12. Sep 3, 2012

### ehild

Which imaginary number?

ehild

13. Sep 4, 2012

### Staff: Mentor

In other words, what symbol do we use to represent $\sqrt{-1}$?

14. Sep 5, 2012

i = sqrt{-1}

15. Sep 5, 2012

### Staff: Mentor

Right

16. Sep 5, 2012

### nebbione

That's right! If x was supposed to be real then, the equation has no solution, but since we are considering complex numbers the equation has a solution!

For the fundamental theorem of Algebra a polynomial form has always a solution in the set of the complex numebers.