Isolating for y? squareoots?

Question

x^2+y^2=12 ..so would you do y^2=12-x^2..so than y=+-sqrt(12-x^2)

but since x^2 and there is a a squareoot over it can you cancel the ^2 out and write it as y=+-sqrt(12)-x

Answer 1

Hey,

First off, you CANNOT do that

sqrt(12-x^2) does NOT equal sqrt(12)-x, when there is a sum or difference inside the sqrt sign you cannot do that, it simply doesn't exist.

your answer is just +/-sqrt(12-x^2), the positive root represents the top half of the circle and the negative root represents the bottom half.

hope this helps

Answer 2

x² + y² = 12

y² = 12 - x²

y = ± √(12 - x²)

------------------------------

y ≠ ± √12 - x

This is the same as saying:

√(25 - 9)

√25 - √9

5 - 3

2

Which is wrong. You always have to do the brackets first so it should be:

√16 = 4