Thread: re-arrange

this is an common algebra error I make. Just want confirm which is the correct answer.




when using the sqrt do I reduce the power on the x also?

In the Real Numbers, you follow something like this stream of logic.

Right-Hand Side
x^4 is positive
8 is positive
1 is positive
Negative sign - The right hand side is negative. It cannot be positive. It cannot be zero.

Left-Hand Side
y^2 is positive

How many positive numbers can you name that are equal to a negative number?

Note: When did 2 get to be the square root of 8? Cube root, perhaps.

It should be obvious that since x can not be 0 and x^4 is always nonnegative, that -1/(8x^4) is always negative. You can not square any number to get a negative number. So you can not solve for y.

Remember that the square root of negative one is imaginary.