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

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

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

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

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

b = 15, b = - 15

Decimal Notation

b = 3.87298 \dots, b = - 3.87298 \dots

Solution steps

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

Move

1^{2}

to the right side

b^{2} = 15

For

x^{2} = f (a)

the solutions are

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

b = 15, b = - 15

z^{2} - 12 z + 36 = 0

x^{2} - x - 56 = 0

x^{2} + 4 x - 21 = 0

0 = x^{2} + 8

2 x^{2} - 8 x + 12 = - 3 x

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