# I Is there any solution to the equation x^(0.5)+1=0?

1. Apr 19, 2017

### Alan Lugano

I got myselft wondering if there is any solution to the equation x^(0.5)+1=0. I know that for a real x there is not but, when I assume x is any imaginary number in the form e^(ix/2) and then solve the equation e^(ix/2)=-1 the result is x=2(2 pi n + pi) for integer values of n. If one then takes n=1, x=6pi and e^(i 6pi) = 1. Whaaaat?

My other question is why in an equation like the one above, squaring both sides to get the solution is not allowed since it would result in x=1.

2. Apr 19, 2017

### Buffu

Why isn't $1$ a solution ?

3. Apr 19, 2017

### BvU

Hello Alan,

You mean 'of the form $e^{ix}$ '
You mean 'and $e^{3\pi} = -1$ '

Question is: is $e^{6\pi}$ in the domain of the function, or is the domain limited to arguments $[0, 2\pi)$ or perhaps $(-\pi , \pi]$ ? Read about it here

(I only get to see the animation when I open the picture in a new tab)

4. Apr 19, 2017

### Alan Lugano

Well I think that once you substitute x=1 in the equation you would get:

1^(0.5) + 1 = 0
2=0 ????

Am I missing something?

5. Apr 19, 2017

### Alan Lugano

Thank you for the answer and the greeting! Let me put it differently:

I have the equation:

$x^{0.5}$ +1 = 0

So I assume that $x=e^{iy}$ and then I get that $e^{iy/2}+1=0$

From the Euler's identity I have a solution in the form $y=2\pi$

So $x=e^{2\pi i}$ which gives $x=1$ and therefore $1^{0.5}+1=0$. Where am I missing something?

6. Apr 19, 2017

### Buffu

No you are correct but also $(-1)^2 = 1$ so $1^{1/2} = \pm 1$. Or $1^{1/2} + 1 = -1 + 1 = 0$.

You were told to find the solutions to the equation.

7. Apr 19, 2017

### Alan Lugano

But if you graph the function $y(x)=1^{x}$ and look at the point $x=0.5$ you will get y=1. In the other hand, if you graph $(-1)^{x}$ you will get $y=i$. So how can you affirm that $1^{1/2}+1=-1+1=0$??

8. Apr 19, 2017

### Buffu

Are you saying that you don't believe that $(-1)^2 = 1$ ?

When you graph a function, the graph will be on the domain and range of the function. For the function $f(x) = \sqrt{x}$, the domain and range is $x\ge 0$ (by convention for range).

You are given to find the solutions to a equation not roots of a function $f(x) = \sqrt{x} + 1$.

9. Apr 19, 2017

### Staff: Mentor

The graph of $y = 1^x$ isn't very interesting, assuming x and y are real numbers. 1 raised to any finite power is just 1, so the graph is just a horizontal line through (0, 1).

Last edited: Apr 19, 2017
10. Apr 19, 2017

### Staff: Mentor

By convention $\sqrt{1} = + 1$, $\sqrt 4 = + 2$, NOT $\pm 1$ or $\pm 2$, respectively. By extension, $1^{1/2} = \sqrt 1$.

11. Apr 19, 2017

### Staff: Mentor

If $z = e^{i\cdot 2\pi}$ = 1 + 0i, then the principal (complex) square root is $e^{i \cdot \pi}$ = -1 + 0i. So this value of z is a solution to the equation $z^{1/2} = -1$, or equivalently $z^{1/2} + 1 = 0$.

I am using z to reinforce the idea that I'm working with complex numbers.

12. Apr 19, 2017

### Alan Lugano

When you wirte $z=-1+0i$ aren't you saying this is a pure real number once the imaginary part is zero? If so, is it right to say that the equation has a real solution. This is very confusing to me.

13. Apr 19, 2017

### Staff: Mentor

z here has no imaginary part, but $z^{1/2}$ is complex, and that is what the equation involves.

14. Apr 19, 2017

### Buffu

Reason for that convention is that to fit $\sqrt{x}$ in functions but that does not matter when solving a equation.

15. Apr 19, 2017

### Staff: Mentor

You are mistaken.
In the real numbers, $\sqrt x = 2$ has one solution -- x = 4.

16. Apr 19, 2017

### pwsnafu

No, it comes from solving equations.

Suppose we are given $a \geq 0$ and we wish to find $x \in \mathbb{R}$ such that $x^2 = a$. Then we show:
1. If $a = 0$ there only exists one solution, namely $x=0$,
2. If $a \neq 0$ then there exists two distinct solutions $x_1$ and $x_2$,
3. We involke the fact that the reals have a total ordering, so that either $x_1 > x_2$ or $x_1 < x_2$.
4. We prove that $x_2 = -x_1$.
After doing all that, we conclude that one of the two solutions is positive and the other is negative. We define the notation $\sqrt{a}$ to be the positive value and call it the principle square root. Due to 4, the other root is then $-\sqrt{a}$.

Further, it's easy to see that "that does not matter when solving a equation" is a false statement by just looking at the quadratic formula. If $\sqrt{a}$ meant both positive and negative the formula would never have the $\pm$ sign in it.

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