Solution to a simple equation

  • #1
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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?
 

Answers and Replies

  • #2
phinds
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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
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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 [tex]\pm 1[/tex]. So what is the solution, I am a bit confused. Would it be

x = -1/2
 
  • #4
NascentOxygen
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Oh I see.... Well, a solution to x^2 could be [tex]\pm 1[/tex]. 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.
 
  • #5
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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
HallsofIvy
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Oh I see.... Well, a solution to x^2 could be [tex]\pm 1[/tex].
No, it's not. You have already said that x^2= -1, not [itex]\pm 1[/itex]. 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.
 
  • #7
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(-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

[tex]x = \sqrt{-1}[/tex]
 
  • #9
Mentallic
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So the solution is

[tex]x = \sqrt{-1}[/tex]

[tex]x=\pm\sqrt{-1}[/tex]
 
  • #10
eumyang
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[tex]x=\pm\sqrt{-1}[/tex]
To the OP: you know the symbol we use for [itex]\sqrt{-1}[/itex], right?
 
  • #11
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Yes, it's the imaginary number. :)
 
  • #12
ehild
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Yes, it's the imaginary number. :)

Which imaginary number?

ehild
 
  • #16
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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.
 

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