What are the solutions to the equation 1^2+b^2=5^2 ?

The solutions to the equation 1^2+b^2=5^2 are b=2sqrt(6),b=-2sqrt(6)

Find the zeros of 1^2+b^2=5^2

The zeros of 1^2+b^2=5^2 are b=2sqrt(6),b=-2sqrt(6)

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1^{2} + b^{2} = 5^{2}

b = 26, b = - 26

Decimal Notation

b = 4.89897 \dots, b = - 4.89897 \dots

Solution steps

1^{2} + b^{2} = 5^{2}

Move

1^{2}

to the right side

b^{2} = 24

For

x^{2} = f (a)

the solutions are

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

b = 24, b = - 24

24 = 26

- 24 = - 26

b = 26, b = - 26

\frac{3 - 5 y}{4} = \frac{2 - 4 y}{3}

n^{3} - 64 n

\frac{x + 1}{2 x - 4} < \frac{1}{\frac{1}{3} x - 1}

\frac{( 8 k ^{- 5} ) ( 2 k ^{3} )}{4 k ^{- 6}}

\frac{3}{- 3}

What are the solutions to the equation 1^2+b^2=5^2 ?
The solutions to the equation 1^2+b^2=5^2 are b=2sqrt(6),b=-2sqrt(6)
Find the zeros of 1^2+b^2=5^2
The zeros of 1^2+b^2=5^2 are b=2sqrt(6),b=-2sqrt(6)