Struggling to Solve Equation: \sqrt{x} + 1 = 0

  • Thread starter m00npirate
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In summary, the equation \sqrt{x} + 1 = 0 has no solution in the real numbers because the right hand side is not in the domain of the left hand side. If we extend the principal square root function to complex numbers, the only possible complex solution would be x = 1. However, by convention, the principal square root of a positive number is taken to be the positive root, so there is no solution to \sqrt{x} = -1. This can also be seen by graphing the function y = \sqrt{x}, where the domain is x >= 0.
  • #1
m00npirate
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I have been trying for the past hour or so to figure out the solution to this equation:

[tex] \sqrt{x} + 1 = 0[/tex]

but to no avail! Here's what i tried...

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

[tex]\sqrt{x} = i^4 [/tex]

[tex]x = \pm i^8 [/tex]

[tex]x = \pm 1[/tex]

But this is most obviously wrong. What did I do wrong and what is the real answer?
 
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  • #2
I think you may be over thinking it, what is the inverse of a square root?
 
  • #3
If I square both sides from the beginning I still get x = +/- 1, neither of which solve the equation.
 
  • #4
You seem to have stated that -1=i4. Check again
 
  • #5
Yea that was a typo, I've figured out what went wrong though. -1 just isn't in the range of the square root function. It has no solution.
 
  • #6
more importantly, x^2=y^2 does not imply x=y.
 
  • #7
[itex]x^2 = y^2 \Rightarrow x = \pm y [/itex]

isn't if and only if?
 
  • #8
more importantly, x^2=y^2 does not imply x=y.

[tex]x^{2} = y^{2} \Rightarrow |x| = |y|[/tex]

Isn't this correct though?Also, MATLAB gave the error 'Warning: Explicit solution could not be found.' when trying to solve the equation.
And Microsoft Math gave x=1, which is clearly incorrect
 
Last edited:
  • #9
marcusmath said:
[tex]x^{2} = y^{2} \Rightarrow |x| = |y|[/tex]

Isn't this correct though?

I said 'x^2=y^2 does not imply x=y'. What you are saying is that x^2=y^2 implies x=+-y. There is no contradiction here.
in the original post, you had

sqrt(x)=-1

then you square both sides and got

x=1 (another mistake in your post was the x=+-1 part; where did the -1 come from??)

and from that you concluded that, if x=1, then

sqrt(x)=-1

which is false, because x^2=y^2 does not imply x=y.There was a post here a while ago, I can't find it, but it expressed a common misconception people make when solving equations. When you have something like x+4=8, and you subtract 4 from both sides to get x=4, you did not prove that

'if x=4 then x+4=8' (1)

you actually proved that

'if x+4=8 then x=4' (2)And (1) is true only if the operation you peformed was injective (i.e. one to one). subtraction is injective, squaring both sides isn't.
 
  • #10
sqrt(x)= -1

Square both side

x=1.

Then

sqrt(1) = 1 or -1

since

1^2 = 1
AND

-1^2 = 1 (*)

Using (*)

sqrt (-1^2) = sqrt (1)

ie -1 = sqrt(1)
 
  • #11
No. [itex]\sqrt{1} = 1[/itex]. By convention, the square root of a number is generally accepted to mean the principal, or positive square root.

So sqrt(x) = -1 has no solution in the reals. If you look at the graph of y = sqrt(x), you'll see that the domain is x >= 0.
 
  • #12
Like Mark44 has said, only the principle root (i.e. the positive number) is taken. This means that there is no complex solution to this equation.
 
  • #13
Mentallic said:
...This means that there is no complex solution to this equation.

I don't yet see why there can be no complex solution. Can this be proven rigorously?
 
  • #14
marcusmath said:
I don't yet see why there can be no complex solution. Can this be proven rigorously?

The right hand side of the equation is not in the domain of the left hand side. If you intend on extending the principal square root function to complex numbers, the problem becomes choosing which complex root is the principal root, as there is no "positive" or "negative" in regards to the complex plane.
 
  • #15
marcusmath said:
I don't yet see why there can be no complex solution. Can this be proven rigorously?

The only complex roots would be [itex]cis(2k\pi)[/itex] for all k integers, which in other words is 1. But since we've already defined by convention that the square root of a positive is only its positive root, then we cannot get the answer [itex]\sqrt{x}=-1[/itex]

If it weren't for this limitation (which some authors don't uphold for convenience in their specific topics of mathematics) then we would have the only possible complex solution x=1
 

1. What does the equation √x + 1 = 0 mean?

The equation √x + 1 = 0 means that the square root of x plus 1 equals zero. In other words, it is asking for what value of x will make the equation true.

2. How do I solve this equation?

To solve this equation, you need to isolate the variable x on one side of the equation. Since the square root and the addition are both operations being performed on x, you can reverse them by subtracting 1 from both sides and then squaring both sides. This will leave you with x = -1, which is your solution.

3. Can this equation have more than one solution?

No, this equation can only have one solution. The square root of a number can only have one positive value, so there is only one possible value of x that will make the equation true.

4. What if I get a negative number when I solve for x?

If you get a negative number when solving for x, then the equation has no solution. The square root of a negative number is undefined in the real number system, so there is no x value that will make the equation true.

5. Can I use a calculator to solve this equation?

Yes, you can use a calculator to solve this equation. However, it is important to understand the steps and concepts behind solving the equation manually before relying on a calculator. Additionally, some calculators may not be able to handle complex numbers, so it is important to check your answer to ensure it is a valid solution.

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