Find All Complex Solutions 121x^-4-483x^-2-4=0

$121 x^{- 4} - 483 x^{- 2} - 4 = 0$

Step 1

Simplify each term.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$121 \frac{1}{x^{4}} - 483 x^{- 2} - 4 = 0$

Combine $121$ and $\frac{1}{x^{4}}$ .

$\frac{121}{x^{4}} - 483 x^{- 2} - 4 = 0$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{121}{x^{4}} - 483 \frac{1}{x^{2}} - 4 = 0$

Combine $- 483$ and $\frac{1}{x^{2}}$ .

$\frac{121}{x^{4}} + \frac{- 483}{x^{2}} - 4 = 0$

Move the negative in front of the fraction.

$\frac{121}{x^{4}} - \frac{483}{x^{2}} - 4 = 0$

Step 2

Find the LCD of the terms in the equation.

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

$x^{4}, x^{2}, 1, 1$

Since $x^{4}, x^{2}, 1, 1$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 1, 1, 1$ then find LCM for the variable part $x^{4}, x^{2}$ .

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factors for $x^{4}$ are $x \cdot x \cdot x \cdot x$ , which is $x$ multiplied by each other $4$ times.

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

$x$ occurs $4$ times.

The factors for $x^{2}$ are $x \cdot x$ , which is $x$ multiplied by each other $2$ times.

$x^{2} = x \cdot x$

$x$ occurs $2$ times.

The LCM of $x^{4}, x^{2}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x \cdot x \cdot x \cdot x$

Simplify $x \cdot x \cdot x \cdot x$ .

Multiply $x$ by $x$ .

$x^{2} \cdot x \cdot x$

Multiply $x^{2}$ by $x$ by adding the exponents.

Multiply $x^{2}$ by $x$ .

Raise $x$ to the power of $1$ .

$x^{2} \cdot x^{1} \cdot x$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{2 + 1} \cdot x$

Add $2$ and $1$ .

$x^{3} \cdot x$

Multiply $x^{3}$ by $x$ by adding the exponents.

Multiply $x^{3}$ by $x$ .

Raise $x$ to the power of $1$ .

$x^{3} \cdot x^{1}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{3 + 1}$

Add $3$ and $1$ .

$x^{4}$

Step 3

Multiply each term in $\frac{121}{x^{4}} - \frac{483}{x^{2}} - 4 = 0$ by $x^{4}$ to eliminate the fractions.

Multiply each term in $\frac{121}{x^{4}} - \frac{483}{x^{2}} - 4 = 0$ by $x^{4}$ .

$\frac{121}{x^{4}} x^{4} - \frac{483}{x^{2}} x^{4} - 4 x^{4} = 0 x^{4}$

Simplify the left side.

Simplify each term.

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

Cancel the common factor.

$\frac{121}{x^{4}} x^{4} - \frac{483}{x^{2}} x^{4} - 4 x^{4} = 0 x^{4}$

Rewrite the expression.

$121 - \frac{483}{x^{2}} x^{4} - 4 x^{4} = 0 x^{4}$

Cancel the common factor of $x^{2}$ .

Move the leading negative in $- \frac{483}{x^{2}}$ into the numerator.

$121 + \frac{- 483}{x^{2}} x^{4} - 4 x^{4} = 0 x^{4}$

Factor $x^{2}$ out of $x^{4}$ .

$121 + \frac{- 483}{x^{2}} (x^{2} x^{2}) - 4 x^{4} = 0 x^{4}$

Cancel the common factor.

$121 + \frac{- 483}{x^{2}} (x^{2} x^{2}) - 4 x^{4} = 0 x^{4}$

Rewrite the expression.

$121 - 483 x^{2} - 4 x^{4} = 0 x^{4}$

Simplify the right side.

Multiply $0$ by $x^{4}$ .

$121 - 483 x^{2} - 4 x^{4} = 0$

Step 4

Solve the equation.

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$- 4 u^{2} - 483 u + 121 = 0$

$u = x^{2}$

Factor the left side of the equation.

Factor $- 1$ out of $- 4 u^{2} - 483 u + 121$ .

Factor $- 1$ out of $- 4 u^{2}$ .

$- (4 u^{2}) - 483 u + 121 = 0$

Factor $- 1$ out of $- 483 u$ .

$- (4 u^{2}) - (483 u) + 121 = 0$

Rewrite $121$ as $- 1 (- 121)$ .

$- (4 u^{2}) - (483 u) - 1 \cdot - 121 = 0$

Factor $- 1$ out of $- (4 u^{2}) - (483 u)$ .

$- (4 u^{2} + 483 u) - 1 \cdot - 121 = 0$

Factor $- 1$ out of $- (4 u^{2} + 483 u) - 1 (- 121)$ .

$- (4 u^{2} + 483 u - 121) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot - 121 = - 484$ and whose sum is $b = 483$ .

Factor $483$ out of $483 u$ .

$- (4 u^{2} + 483 (u) - 121) = 0$

Rewrite $483$ as $- 1$ plus $484$

$- (4 u^{2} + (- 1 + 484) u - 121) = 0$

Apply the distributive property.

$- (4 u^{2} - 1 u + 484 u - 121) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((4 u^{2} - 1 u) + 484 u - 121) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (u (4 u - 1) + 121 (4 u - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$- ((4 u - 1) (u + 121)) = 0$

Remove unnecessary parentheses.

$- (4 u - 1) (u + 121) = 0$

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

$4 u - 1 = 0$

$u + 121 = 0$

Set $4 u - 1$ equal to $0$ and solve for $u$ .

Set $4 u - 1$ equal to $0$ .

$4 u - 1 = 0$

Solve $4 u - 1 = 0$ for $u$ .

Add $1$ to both sides of the equation.

$4 u = 1$

Divide each term in $4 u = 1$ by $4$ and simplify.

Divide each term in $4 u = 1$ by $4$ .

$\frac{4 u}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 u}{4} = \frac{1}{4}$

Divide $u$ by $1$ .

$u = \frac{1}{4}$

Set $u + 121$ equal to $0$ and solve for $u$ .

Set $u + 121$ equal to $0$ .

$u + 121 = 0$

Subtract $121$ from both sides of the equation.

$u = - 121$

The final solution is all the values that make $- (4 u - 1) (u + 121) = 0$ true.

$u = \frac{1}{4}, - 121$

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = \frac{1}{4}$

${(x^{2})}^{1} = - 121$

Solve the first equation for $x$ .

$x^{2} = \frac{1}{4}$

Solve the equation for $x$ .

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

$x = \pm \sqrt{\frac{1}{4}}$

Simplify $\pm \sqrt{\frac{1}{4}}$ .

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{4}}$

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{4}}$

Simplify the denominator.

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

$x = \pm \frac{1}{\sqrt{2^{2}}}$

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

$x = \pm \frac{1}{2}$

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.

$x = \frac{1}{2}$

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

$x = - \frac{1}{2}$

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

$x = \frac{1}{2}, - \frac{1}{2}$

Solve the second equation for $x$ .

${(x^{2})}^{1} = - 121$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = - 121$

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

$x = \pm \sqrt{- 121}$

Simplify $\pm \sqrt{- 121}$ .

Rewrite $- 121$ as $- 1 (121)$ .

$x = \pm \sqrt{- 1 (121)}$

Rewrite $\sqrt{- 1 (121)}$ as $\sqrt{- 1} \cdot \sqrt{121}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{121}$

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

$x = \pm i \cdot \sqrt{121}$

Rewrite $121$ as $11^{2}$ .

$x = \pm i \cdot \sqrt{11^{2}}$

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

$x = \pm i \cdot 11$

Move $11$ to the left of $i$ .

$x = \pm 11 i$

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.

$x = 11 i$

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

$x = - 11 i$

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

$x = 11 i, - 11 i$

The solution to $- 4 x^{4} - 483 x^{2} + 121 = 0$ is $x = \frac{1}{2}, - \frac{1}{2}, 11 i, - 11 i$ .

$x = \frac{1}{2}, - \frac{1}{2}, 11 i, - 11 i$

Find All Complex Solutions 121x^-4-483x^-2-4=0

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