Is x equal to the 4th root of y in the equation y=x^4?

[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!?

 

[disregardthat]

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

 

[Mark44]

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!?

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|.

 

[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.

 

[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.

 

[Mark44]

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

Why do this? The absolute value of a real number is nonnegative, so there's no point in taking the absolute value again.

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

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.

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 $f^{-1}(x) = \pm \sqrt[4]{x}$.

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 $\pm$.

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.