Solve for a a^2+b^2=10^2

$a^{2} + b^{2} = 10^{2}$

Raise $10$ to the power of $2$ .

$a^{2} + b^{2} = 100$

Subtract $b^{2}$ from both sides of the equation.

$a^{2} = 100 - b^{2}$

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

$a = \pm \sqrt{100 - b^{2}}$

Simplify $\pm \sqrt{100 - b^{2}}$ .

Rewrite $100$ as $10^{2}$ .

$a = \pm \sqrt{10^{2} - b^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 10$ and $b = b$ .

$a = \pm \sqrt{(10 + b) (10 - b)}$

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.

$a = \sqrt{(10 + b) (10 - b)}$

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

$a = - \sqrt{(10 + b) (10 - b)}$

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

$a = \sqrt{(10 + b) (10 - b)}$

$a = - \sqrt{(10 + b) (10 - b)}$

$a = \sqrt{(10 + b) (10 - b)}$

$a = - \sqrt{(10 + b) (10 - b)}$