What are the solutions to the equation 1^2+b^2=3^2 ?

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

Find the zeros of 1^2+b^2=3^2

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

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1^{2} + b^{2} = 3^{2}

b = 22, b = - 22

Decimal Notation

b = 2.82842 \dots, b = - 2.82842 \dots

Solution steps

1^{2} + b^{2} = 3^{2}

Move

1^{2}

to the right side

b^{2} = 8

For

x^{2} = f (a)

the solutions are

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

b = 8, b = - 8

8 = 22

- 8 = - 22

b = 22, b = - 22

n^{2} - 3 n = - 7

- 2 x^{2} + x + 1

- x = 1

\frac{x}{9} = 81

\frac{1}{4} (x - 4) + \frac{2}{5} (x - 5) = \frac{1}{5} (x^{2} - 53)

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