Solve for y y^2=x^2-1

$y^{2} = x^{2} - 1$

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

$y = \pm \sqrt{x^{2} - 1}$

Simplify $\pm \sqrt{x^{2} - 1}$ .

Rewrite $1$ as $1^{2}$ .

$y = \pm \sqrt{x^{2} - 1^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$y = \pm \sqrt{(x + 1) (x - 1)}$

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.

$y = \sqrt{(x + 1) (x - 1)}$

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

$y = - \sqrt{(x + 1) (x - 1)}$

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

$y = \sqrt{(x + 1) (x - 1)}$

$y = - \sqrt{(x + 1) (x - 1)}$

$y = \sqrt{(x + 1) (x - 1)}$

$y = - \sqrt{(x + 1) (x - 1)}$