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

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In the equation y=x^4, x is equal to ±√[4]{y}, as this represents the real solutions when y is non-negative. The discussion clarifies that while there are two real fourth roots for positive y, if y is not positive, all roots are complex. The notation used for absolute values and norms is debated, with suggestions that simpler expressions would suffice. The inverse function of f(x) = x^4 is correctly identified as f^{-1}(x) = √[4]{x} when the domain is restricted to non-negative x. Overall, the conversation emphasizes the importance of clarity and simplicity in mathematical expressions.
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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!?
 
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If y is non-negative, this is indeed the real solutions for x.
 
Jhenrique said:
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|.
 
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.
 
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.
 
Jhenrique said:
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.
Jhenrique said:
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.
Jhenrique said:
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##.
Jhenrique said:
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.
 
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