What are the solutions to the equation 1^2+4^2=c^2 ?

The solutions to the equation 1^2+4^2=c^2 are c=sqrt(17),c=-sqrt(17)

Find the zeros of 1^2+4^2=c^2

The zeros of 1^2+4^2=c^2 are c=sqrt(17),c=-sqrt(17)

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1^{2} + 4^{2} = c^{2}

c = 17, c = - 17

Solution steps

1^{2} + 4^{2} = c^{2}

Switch sides

c^{2} = 1^{2} + 4^{2}

Simplify

1^{2} + 4^{2} : 17

For

x^{2} = f (a)

the solutions are

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

c = 17, c = - 17

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(4 x y^{4})^{3}

\frac{( x + 3 ) ( x - 2 )}{( x + 7 ) ^{2}} \geq 0

2048

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What are the solutions to the equation 1^2+4^2=c^2 ?
The solutions to the equation 1^2+4^2=c^2 are c=sqrt(17),c=-sqrt(17)
Find the zeros of 1^2+4^2=c^2
The zeros of 1^2+4^2=c^2 are c=sqrt(17),c=-sqrt(17)