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

The solutions to the equation 1^2+b^2=2^2 are b=sqrt(3),b=-sqrt(3)

Find the zeros of 1^2+b^2=2^2

The zeros of 1^2+b^2=2^2 are b=sqrt(3),b=-sqrt(3)

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1^{2} + b^{2} = 2^{2}

b = 3, b = - 3

Decimal Notation

b = 1.73205 \dots, b = - 1.73205 \dots

Solution steps

1^{2} + b^{2} = 2^{2}

Move

1^{2}

to the right side

b^{2} = 3

For

x^{2} = f (a)

the solutions are

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

b = 3, b = - 3

x - 3 = 2 + x

3 x^{2} - 16 x + 6 = 0

\frac{1}{2} - \frac{1}{8}

\frac{1}{x - 1} + \frac{2}{x} < 0

\frac{p + 1}{2} = \frac{3 p + 7}{5}

What are the solutions to the equation 1^2+b^2=2^2 ?
The solutions to the equation 1^2+b^2=2^2 are b=sqrt(3),b=-sqrt(3)
Find the zeros of 1^2+b^2=2^2
The zeros of 1^2+b^2=2^2 are b=sqrt(3),b=-sqrt(3)