Find All Complex Solutions 1/3x^4=27

$\frac{1}{3} x^{4} = 27$

Step 1

Multiply each term in $\frac{1}{3} x^{4} = 27$ by $3$ to eliminate the fractions.

Multiply each term in $\frac{1}{3} x^{4} = 27$ by $3$ .

$\frac{1}{3} x^{4} \cdot 3 = 27 \cdot 3$

Simplify the left side.

Combine $\frac{1}{3}$ and $x^{4}$ .

$\frac{x^{4}}{3} \cdot 3 = 27 \cdot 3$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{x^{4}}{3} \cdot 3 = 27 \cdot 3$

Rewrite the expression.

$x^{4} = 27 \cdot 3$

Simplify the right side.

Multiply $27$ by $3$ .

$x^{4} = 81$

Step 2

Subtract $81$ from both sides of the equation.

$x^{4} - 81 = 0$

Step 3

Factor the left side of the equation.

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

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

Rewrite $81$ as $9^{2}$ .

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

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 $b = 9$ .

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

Simplify.

Rewrite $9$ as $3^{2}$ .

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

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 $b = 3$ .

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

Remove unnecessary parentheses.

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

Step 4

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

$x^{2} + 9 = 0$

$x + 3 = 0$

$x - 3 = 0$

Step 5

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

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

$x^{2} + 9 = 0$

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

Subtract $9$ from both sides of the equation.

$x^{2} = - 9$

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

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

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

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

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

$x = \pm i \cdot 3$

Move $3$ to the left of $i$ .

$x = \pm 3 i$

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.

$x = 3 i$

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

$x = - 3 i$

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

$x = 3 i, - 3 i$

Step 6

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 8

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

$x = 3 i, - 3 i, - 3, 3$

Find All Complex Solutions 1/3x^4=27

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