Find All Complex Solutions 81x^4-16=0

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

Step 1

Factor the left side of the equation.

Rewrite $81 x^{4}$ as ${(9 x^{2})}^{2}$ .

${(9 x^{2})}^{2} - 16 = 0$

Rewrite $16$ as $4^{2}$ .

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

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

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

Simplify.

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

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

Rewrite $4$ as $2^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 2$ .

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

Remove unnecessary parentheses.

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

Step 2

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

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

$3 x + 2 = 0$

$3 x - 2 = 0$

Step 3

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

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

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

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

Subtract $4$ from both sides of the equation.

$9 x^{2} = - 4$

Divide each term in $9 x^{2} = - 4$ by $9$ and simplify.

Divide each term in $9 x^{2} = - 4$ by $9$ .

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

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Move the negative in front of the fraction.

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

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 4

Set $3 x + 2$ equal to $0$ and solve for $x$ .

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Solve $3 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term in $3 x = - 2$ by $3$ and simplify.

Divide each term in $3 x = - 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 5

Set $3 x - 2$ equal to $0$ and solve for $x$ .

Set $3 x - 2$ equal to $0$ .

$3 x - 2 = 0$

Solve $3 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 6

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

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

Find All Complex Solutions 81x^4-16=0

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