Solve for x 15x^2=-x^2-1

$15 x^{2} = - x^{2} - 1$

Step 1

Move all terms containing $x$ to the left side of the equation.

Add $x^{2}$ to both sides of the equation.

$15 x^{2} + x^{2} = - 1$

Add $15 x^{2}$ and $x^{2}$ .

$16 x^{2} = - 1$

Step 2

Divide each term in $16 x^{2} = - 1$ by $16$ and simplify.

Divide each term in $16 x^{2} = - 1$ by $16$ .

$\frac{16 x^{2}}{16} = \frac{- 1}{16}$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 x^{2}}{16} = \frac{- 1}{16}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 1}{16}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} = - \frac{1}{16}$

Step 3

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

$x = \pm \sqrt{- \frac{1}{16}}$

Step 4

Simplify $\pm \sqrt{- \frac{1}{16}}$ .

Rewrite $- \frac{1}{16}$ as ${(\frac{i}{4})}^{2}$ .

$x = \pm \sqrt{{(\frac{i}{4})}^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm \frac{i}{4}$

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.

$x = \frac{i}{4}$

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

$x = - \frac{i}{4}$

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

$x = \frac{i}{4}, - \frac{i}{4}$