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

The solutions to the equation 1^2+x^2=4^2 are x=sqrt(15),x=-sqrt(15)

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

The zeros of 1^2+x^2=4^2 are x=sqrt(15),x=-sqrt(15)

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

x = 15, x = - 15

Decimal Notation

x = 3.87298 \dots, x = - 3.87298 \dots

Solution steps

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

Move

1^{2}

to the right side

x^{2} = 15

For

x^{2} = f (a)

the solutions are

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

x = 15, x = - 15

7 x^{2} = 6 x

\frac{5 ( 7 - z )}{4} = - z

\frac{1}{x - 2} + \frac{5}{2 x - 18} \geq 0

8 - \frac{x}{2} > 6

What are the solutions to the equation 1^2+x^2=4^2 ?
The solutions to the equation 1^2+x^2=4^2 are x=sqrt(15),x=-sqrt(15)
Find the zeros of 1^2+x^2=4^2
The zeros of 1^2+x^2=4^2 are x=sqrt(15),x=-sqrt(15)