What are the solutions to the equation 40^2+b^2=41^2 ?

The solutions to the equation 40^2+b^2=41^2 are b=9,b=-9

Find the zeros of 40^2+b^2=41^2

The zeros of 40^2+b^2=41^2 are b=9,b=-9

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4 0^{2} + b^{2} = 4 1^{2}

b = 9, b = - 9

Solution steps

4 0^{2} + b^{2} = 4 1^{2}

Move

4 0^{2}

to the right side

b^{2} = 81

For

x^{2} = f (a)

the solutions are

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

b = 81, b = - 81

81 = 9

- 81 = - 9

b = 9, b = - 9

(\frac{2}{5})^{2} \cdot 25

3 h^{2} + 11 h + 6

x^{2} + 2 x - 12 \leq 0

\frac{1}{2} (6 x + 8) = 2 x + 4 + x

(x^{\frac{1}{3}})^{\frac{1}{3}}

What are the solutions to the equation 40^2+b^2=41^2 ?
The solutions to the equation 40^2+b^2=41^2 are b=9,b=-9
Find the zeros of 40^2+b^2=41^2
The zeros of 40^2+b^2=41^2 are b=9,b=-9