# Root quartic

1. Apr 14, 2014

### Jhenrique

If $y=x^4$, so x is equal to $\pm \sqrt[4]{y}$ or x is equal to $\pm \sqrt[2]{\pm \sqrt[2]{y}}$ ?

Well, I think that x is qual to $\pm \sqrt[4]{y}$ because

$\\y=x^4 \\\sqrt[4]{y} = \sqrt[4]{x^4} = \sqrt[2]{\sqrt[2]{(x^2)^2}} = \sqrt[2]{|x^2|} = \sqrt[2]{|x|^2} = ||x|| = |x| \\ \pm \sqrt[4]{y}=x$

Right!?

2. Apr 14, 2014

### disregardthat

If y is non-negative, this is indeed the real solutions for x.

3. Apr 14, 2014

### Staff: Mentor

This is a basic algebra question that you should be able to answer for yourself.

If y = 16 = 24, then y = $\pm$2. These are the only real fourth roots of 16. The other two roots are imaginary.

Some of what you wrote above is unnecessary, such as replacing x2 by |x2|. If x is real, then x2 ≥ 0, so x2 and |x2| represent the same number.

This notation -- ||x|| -- means the norm of x, in which the context is usually that x is a vector or something in a vector space. It's overkill to use ||x|| when all you mean is the absolute value, |x|.

4. Apr 14, 2014

### HallsofIvy

Note that people are saying "the real roots". If y is a positive real number then it has 2 real fourth roots and two imaginary roots. If y is not a positive real number then all four fourth roots are complex numbers.

5. Apr 15, 2014

### Jhenrique

A) ||x|| isn't the norm of, is the abs of abs of x.

B) I didn't omit the abs in |x|² for efect of step-by-step.

C) All roots of equation $y = x^4$ is given by $x=\pm\sqrt[2]{\pm\sqrt[2]{y}}$. But, if $f(x) = x^4$, the inverse function is given simply $f^{-1}(x) = \pm \sqrt[4]{x}$. Also, if $f(z) = z^4$, thus maybe the inverse function is probably given by $f^{-1}(z) = \pm\sqrt[2]{\pm\sqrt[2]{z}}$ and not by $f^{-1}(z) = \pm \sqrt[4]{z}$. But, I don't know how to verify this, cause I don't know how and where I can plot a complex graphic.

6. Apr 15, 2014

### Staff: Mentor

Why do this? The absolute value of a real number is nonnegative, so there's no point in taking the absolute value again.
Then you're adding extra, unnecessary steps. All you need to say is that $\sqrt{x^2} = |x|$. And as I mentioned in another thread, including the index of 2 on your square root is completely unnecessary. This is one of a number of shortcuts that we take in mathematics. For example, we rarely write 1x in place of x, or y1 when we mean y.
No.
What you wrote for the inverse is not a function. f is not a one-to-one function, so it doesn't have an inverse. However, if we restrict the domain to x ≥ 0, then f is now one-to-one, and its inverse is f-1(x) = $\sqrt[4]{x}$. No $\pm$.