Ok I'm a little confused with this: 16x-y^4=0

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The discussion centers on the equation 16x - y^4 = 0 and the implications of solving for y. When isolating y, the solution is y = ±4th root(16x), indicating that it is not a function due to the dual outputs. The conversation also addresses the square root function, specifically y = √(1 - x), clarifying that the square root is defined as the positive root, but when solving for y, both the positive and negative roots must be considered. Participants conclude that the presence of a plus/minus sign is necessary only when explicitly solving for y.

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Ok I'm a little confused with this:

16x-y^4=0

When you solve for (y) you get: y= plus/minus 4th root(16x) right?
There for it isn't a function.

What about y=sq.rt.(1-x)? Does it have the plus/minus automatically? Do you only put the plus/minus sign if you solve for y?

I just don't get when you have to put the plus/minus in front of the sq.rt. Do you always have to put one there? or only if you solve for y, etc.?
 
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Well, the even root of a number is DEFINED to be the positive number.
That is, by definition, \sqrt{1-x} is a positive number.

When you solve for y, however, you must remember that not only do you have the positive, even root as a solution, but also the negative of that number.
 
Oh ok. So if they give you an equation that has no plus/minus infront of it, then the equation is POSITIVE. However, if you solve for y and get a sq.rt. then you must put the plus/minus sign right? Thanks.
 

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