Find All Complex Solutions x^4=16/81

$x^{4} = \frac{16}{81}$

Step 1

Subtract $\frac{16}{81}$ from both sides of the equation.

$x^{4} - \frac{16}{81} = 0$

Step 2

Factor the left side of the equation.

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

${(x^{2})}^{2} - \frac{16}{81} = 0$

Rewrite $\frac{16}{81}$ as ${(\frac{4}{9})}^{2}$ .

${(x^{2})}^{2} - {(\frac{4}{9})}^{2} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = \frac{4}{9}$ .

$(x^{2} + \frac{4}{9}) (x^{2} - \frac{4}{9}) = 0$

Simplify.

Rewrite $\frac{4}{9}$ as ${(\frac{2}{3})}^{2}$ .

$(x^{2} + \frac{4}{9}) (x^{2} - {(\frac{2}{3})}^{2}) = 0$

Factor.

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 $b = \frac{2}{3}$ .

$(x^{2} + \frac{4}{9}) ((x + \frac{2}{3}) (x - \frac{2}{3})) = 0$

Remove unnecessary parentheses.

$(x^{2} + \frac{4}{9}) (x + \frac{2}{3}) (x - \frac{2}{3}) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} + \frac{4}{9} = 0$

$x + \frac{2}{3} = 0$

$x - \frac{2}{3} = 0$

Step 4

Set $x^{2} + \frac{4}{9}$ equal to $0$ and solve for $x$ .

Set $x^{2} + \frac{4}{9}$ equal to $0$ .

$x^{2} + \frac{4}{9} = 0$

Solve $x^{2} + \frac{4}{9} = 0$ for $x$ .

Subtract $\frac{4}{9}$ from both sides of the equation.

$x^{2} = - \frac{4}{9}$

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

$x = \pm \sqrt{- \frac{4}{9}}$

Simplify $\pm \sqrt{- \frac{4}{9}}$ .

Rewrite $- \frac{4}{9}$ as ${(\frac{2 i}{3})}^{2}$ .

$x = \pm \sqrt{{(\frac{2 i}{3})}^{2}}$

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

$x = \pm \frac{2 i}{3}$

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.

$x = \frac{2 i}{3}$

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

$x = - \frac{2 i}{3}$

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

$x = \frac{2 i}{3}, - \frac{2 i}{3}$

Step 5

Set $x + \frac{2}{3}$ equal to $0$ and solve for $x$ .

Set $x + \frac{2}{3}$ equal to $0$ .

$x + \frac{2}{3} = 0$

Subtract $\frac{2}{3}$ from both sides of the equation.

$x = - \frac{2}{3}$

Step 6

Set $x - \frac{2}{3}$ equal to $0$ and solve for $x$ .

Set $x - \frac{2}{3}$ equal to $0$ .

$x - \frac{2}{3} = 0$

Add $\frac{2}{3}$ to both sides of the equation.

$x = \frac{2}{3}$

Step 7

The final solution is all the values that make $(x^{2} + \frac{4}{9}) (x + \frac{2}{3}) (x - \frac{2}{3}) = 0$ true.

$x = \frac{2 i}{3}, - \frac{2 i}{3}, - \frac{2}{3}, \frac{2}{3}$

Find All Complex Solutions x^4=16/81

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