Why is x^2 = -1 not a possible solution for x in the equation x^2 + 1 = 0?

[help1please]

Homework Statement

X^2 + 1 = 0

Homework Equations

find the solution

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?

 

[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"

 

[help1please]

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"

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

 

[NascentOxygen]

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

(-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.

 

[micromass]

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.

 

[HallsofIvy]

Oh I see... Well, a solution to x^2 could be $$\pm 1$$.

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?

So what is the solution, I am a bit confused. Would it be

x = -1/2

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.

 

[help1please]

(-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.

So the solution is

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

 

[phinds]

So the solution is

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

Bravo.

 

[Mentallic]

So the solution is

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

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

 

[eumyang]

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

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

 

[help1please]

Yes, it's the imaginary number. :)

 

[ehild]

Yes, it's the imaginary number. :)

Which imaginary number?

ehild

 

[Mark44]

Yes, it's the imaginary number. :)

Which imaginary number?

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

 

[help1please]

i = sqrt{-1}

 

[Mark44]

i = sqrt{-1}

Right

 

[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.