What is the answer to y=sqrt(36-y^2),0<= y<= 3 ?

The answer to y=sqrt(36-y^2),0<= y<= 3 is y=3sqrt(2)

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y=sqrt(36-y^2),0<= y<= 3

The answer to y=sqrt(36-y^2),0<= y<= 3 is y=3sqrt(2)

y = 36 - y^{2}, 0 \leq y \leq 3

y = 32

Solution steps

y = 36 - y^{2}

Square both sides:

y^{2} = 36 - y^{2}

y^{2} = 36 - y^{2}

Solve

y^{2} = 36 - y^{2} : y = 32, y = - 32

y = 32, y = - 32

Verify Solutions:

y = 32

True

, y = - 32

False

The solution is

y = 32

Check constrains

y = 32

2 y - 3 = 3 y^{2} - 10 y + 12

14 x - 6 - 3 = x

x, v = 2 a x

6 x + 64 = x - 16

x^{2} + 8 x + 6 = x^{2} + 8 x

What is the answer to y=sqrt(36-y^2),0<= y<= 3 ?
The answer to y=sqrt(36-y^2),0<= y<= 3 is y=3sqrt(2)
Solve for y: y=sqrt(36-y^2),0<= y<= 3
The solution is y=3sqrt(2)