Solve for b a^2+b^2=c^2 ; for b

$a^{2} + b^{2} = c^{2}$ ; for $b$

Step 1

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

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

Step 2

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

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

Step 3

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

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

Step 4

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.

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

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

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

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

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

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

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

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