What is the answer to solvefor y,x=sqrt(4+y^2) ?

The answer to solvefor y,x=sqrt(4+y^2) is y=sqrt(x^2-4) {x>= 2},y=-sqrt(x^2-4) {x>= 2}

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solvefor y,x=sqrt(4+y^2)

solve for

y, x = 4 + y^{2}

y = x^{2} - 4 {x \geq 2}, y = - x^{2} - 4 {x \geq 2}

Solution steps

x = 4 + y^{2}

Square both sides:

x^{2} = 4 + y^{2}

x^{2} = 4 + y^{2}

Solve

x^{2} = 4 + y^{2} : y = x^{2} - 4, y = - x^{2} - 4

y = x^{2} - 4, y = - x^{2} - 4

Verify Solutions:

y = x^{2} - 4 {x \geq 2}, y = - x^{2} - 4 {x \geq 2}

The solutions are

y = x^{2} - 4 {x \geq 2}, y = - x^{2} - 4 {x \geq 2}

x, 50 = 100 (x)^{24.1}

x + 17 = x + 5

x, 3 x - 1 = x^{2} - x

3 (x + 9) - 3 = x

x + 18 - x - 10 = 2

What is the answer to solvefor y,x=sqrt(4+y^2) ?
The answer to solvefor y,x=sqrt(4+y^2) is y=sqrt(x^2-4) {x>= 2},y=-sqrt(x^2-4) {x>= 2}