Solve by Factoring 16x^4-81=0

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

Rewrite $16 x^{4}$ as ${(4 x^{2})}^{2}$ .

${(4 x^{2})}^{2} - 81 = 0$

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

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

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

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

Simplify.

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

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

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

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

Factor.

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

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

Remove unnecessary parentheses.

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

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

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

$2 x + 3 = 0$

$2 x - 3 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

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

Subtract $9$ from both sides of the equation.

$4 x^{2} = - 9$

Divide each term by $4$ and simplify.

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

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

Cancel the common factor of $4$ .

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Cancel the common factor.

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

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

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

Move the negative in front of the fraction.

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

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

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

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

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Simplify the right side of the equation.

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

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

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

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

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

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First, use the positive value of the $\pm$ to find the first solution.

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

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

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

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

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

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$2 x + 3 = 0$

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term by $2$ and simplify.

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

Move the negative in front of the fraction.

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

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$2 x - 3 = 0$

Add $3$ to both sides of the equation.

$2 x = 3$

Divide each term by $2$ and simplify.

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Solve by Factoring 16x^4-81=0

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