Precalculus Examples

$1 = \frac{18 x^{2}}{x^{4} + 81}$

Step 1

Rewrite the equation as $\frac{18 x^{2}}{x^{4} + 81} = 1$ .

$\frac{18 x^{2}}{x^{4} + 81} = 1$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x^{4} + 81, 1$

Step 2.2

Remove parentheses.

$x^{4} + 81, 1$

Step 2.3

The LCM of one and any expression is the expression.

$x^{4} + 81$

Step 3

Multiply each term in $\frac{18 x^{2}}{x^{4} + 81} = 1$ by $x^{4} + 81$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{18 x^{2}}{x^{4} + 81} = 1$ by $x^{4} + 81$ .

$\frac{18 x^{2}}{x^{4} + 81} (x^{4} + 81) = 1 (x^{4} + 81)$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $x^{4} + 81$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{18 x^{2}}{x^{4} + 81} (x^{4} + 81) = 1 (x^{4} + 81)$

Step 3.2.1.2

Rewrite the expression.

$18 x^{2} = 1 (x^{4} + 81)$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply $x^{4} + 81$ by $1$ .

$18 x^{2} = x^{4} + 81$

Step 4

Solve the equation.

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Step 4.1

Subtract $x^{4}$ from both sides of the equation.

$18 x^{2} - x^{4} = 81$

Step 4.2

Subtract $81$ from both sides of the equation.

$18 x^{2} - x^{4} - 81 = 0$

Step 4.3

Factor the left side of the equation.

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Step 4.3.1

Factor $- 1$ out of $18 x^{2} - x^{4} - 81$ .

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Step 4.3.1.1

Reorder $18 x^{2}$ and $- x^{4}$ .

$- x^{4} + 18 x^{2} - 81 = 0$

Step 4.3.1.2

Factor $- 1$ out of $- x^{4}$ .

$- (x^{4}) + 18 x^{2} - 81 = 0$

Step 4.3.1.3

Factor $- 1$ out of $18 x^{2}$ .

$- (x^{4}) - (- 18 x^{2}) - 81 = 0$

Step 4.3.1.4

Rewrite $- 81$ as $- 1 (81)$ .

$- (x^{4}) - (- 18 x^{2}) - 1 \cdot 81 = 0$

Step 4.3.1.5

Factor $- 1$ out of $- (x^{4}) - (- 18 x^{2})$ .

$- (x^{4} - 18 x^{2}) - 1 \cdot 81 = 0$

Step 4.3.1.6

Factor $- 1$ out of $- (x^{4} - 18 x^{2}) - 1 (81)$ .

$- (x^{4} - 18 x^{2} + 81) = 0$

Step 4.3.2

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

$- ({(x^{2})}^{2} - 18 x^{2} + 81) = 0$

Step 4.3.3

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- (u^{2} - 18 u + 81) = 0$

Step 4.3.4

Factor using the perfect square rule.

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Step 4.3.4.1

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

$- (u^{2} - 18 u + 9^{2}) = 0$

Step 4.3.4.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$18 u = 2 \cdot u \cdot 9$

Step 4.3.4.3

Rewrite the polynomial.

$- (u^{2} - 2 \cdot u \cdot 9 + 9^{2}) = 0$

Step 4.3.4.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 9$ .

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

Step 4.3.5

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 4.3.6

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

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

Step 4.3.7

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 $b = 3$ .

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

Step 4.3.8

Factor.

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Step 4.3.8.1

Apply the product rule to $(x + 3) (x - 3)$ .

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

Step 4.3.8.2

Remove unnecessary parentheses.

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

Step 4.4

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

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

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

Step 4.5

Set ${(x + 3)}^{2}$ equal to $0$ and solve for $x$ .

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Step 4.5.1

Set ${(x + 3)}^{2}$ equal to $0$ .

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

Step 4.5.2

Solve ${(x + 3)}^{2} = 0$ for $x$ .

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Step 4.5.2.1

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

$x + 3 = 0$

Step 4.5.2.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 4.6

Set ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

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Step 4.6.1

Set ${(x - 3)}^{2}$ equal to $0$ .

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

Step 4.6.2

Solve ${(x - 3)}^{2} = 0$ for $x$ .

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Step 4.6.2.1

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

$x - 3 = 0$

Step 4.6.2.2

Add $3$ to both sides of the equation.

$x = 3$

Step 4.7

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

$x = - 3, 3$