What is the answer to (sqrt(1-y^2))^2+y^2-1=0 ?

The answer to (sqrt(1-y^2))^2+y^2-1=0 is -1<= y<= 1

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(sqrt(1-y^2))^2+y^2-1=0

(1 - y^{2})^{2} + y^{2} - 1 = 0

- 1 \leq y \leq 1

Solution steps

(1 - y^{2})^{2} + y^{2} - 1 = 0

Expand

(1 - y^{2})^{2} + y^{2} - 1 : 0

0 = 0

Both sides are equal

True for all y

Verify Solutions:

y = - 1

True

, - 1 < y < 1

True

, y = 1

True

The solution is

- 1 \leq y \leq 1

13 y + 9 - y + 9 = 6

5 + x = 8

y, x = y p - 1

x - x = 1

x + 6 + x - 2 - 4 = 0

What is the answer to (sqrt(1-y^2))^2+y^2-1=0 ?
The answer to (sqrt(1-y^2))^2+y^2-1=0 is -1<= y<= 1
Solve for y: (sqrt(1-y^2))^2+y^2-1=0
The solution is -1<= y<= 1