hi, i have a question regarding what is the correct way of doing this problem
given that 5 - (y^2)^(1/2) = -3, what is x?
the y term is actually squared and underneath a radical sign. i thought you could solve it as:
5 - y = -3
y = 8
i thought that the y^2^(1/2) would cancel each other out, but the book says that you should do it this way:
5+3 = (y^2)^(1/2)
64 = y^2
y = +8, -8
i actually also think there is a third way of doing it:
25 - y^2 = 9
y^2 = 16
y = +4, -4
but when plugging 4 back into the original, 4 doesnt work, but the math sure works when solving for 4, could someone elaborate?
thank you
given that 5 - (y^2)^(1/2) = -3, what is x?
the y term is actually squared and underneath a radical sign. i thought you could solve it as:
5 - y = -3
y = 8
i thought that the y^2^(1/2) would cancel each other out, but the book says that you should do it this way:
5+3 = (y^2)^(1/2)
64 = y^2
y = +8, -8
i actually also think there is a third way of doing it:
25 - y^2 = 9
y^2 = 16
y = +4, -4
but when plugging 4 back into the original, 4 doesnt work, but the math sure works when solving for 4, could someone elaborate?
thank you
