What are the solutions to the equation solvefor k,p^2+q^2=k^2 ?

The solutions to the equation solvefor k,p^2+q^2=k^2 are k=sqrt(p^2+q^2),k=-sqrt(p^2+q^2)

Find the zeros of solvefor k,p^2+q^2=k^2

The zeros of solvefor k,p^2+q^2=k^2 are k=sqrt(p^2+q^2),k=-sqrt(p^2+q^2)

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solve for

k, p^{2} + q^{2} = k^{2}

k = p^{2} + q^{2}, k = - p^{2} + q^{2}

Solution steps

p^{2} + q^{2} = k^{2}

Switch sides

k^{2} = p^{2} + q^{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

k = p^{2} + q^{2}, k = - p^{2} + q^{2}

(k - 12) \cdot x^{2} + 2 (k - 12) \cdot x + 2 = 0

2 \cdot x^{2} + 7 x + 52 = 0

x, x^{2} \cdot p^{2} + y^{2} \cdot q^{2} = z^{2}

(x + 2)^{3} = x^{3} - 8

lo g_{10} (12) (x^{2} - x) = 1

What are the solutions to the equation solvefor k,p^2+q^2=k^2 ?
The solutions to the equation solvefor k,p^2+q^2=k^2 are k=sqrt(p^2+q^2),k=-sqrt(p^2+q^2)
Find the zeros of solvefor k,p^2+q^2=k^2
The zeros of solvefor k,p^2+q^2=k^2 are k=sqrt(p^2+q^2),k=-sqrt(p^2+q^2)