Solve for b 1^2+b^2=4^2

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

Step 1

One to any power is one.

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

Step 2

Raise $4$ to the power of $2$ .

$1 + b^{2} = 16$

Step 3

Move all terms not containing $b$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$b^{2} = 16 - 1$

Subtract $1$ from $16$ .

$b^{2} = 15$

Step 4

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt{15}$

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$b = \sqrt{15}$

Next, use the negative value of the $\pm$ to find the second solution.

$b = - \sqrt{15}$

The complete solution is the result of both the positive and negative portions of the solution.

$b = \sqrt{15}, - \sqrt{15}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$b = \sqrt{15}, - \sqrt{15}$

Decimal Form:

$b = 3.87298334 \dots, - 3.87298334 \dots$