Solve for x 4/((x+1)^2)-1/((x-1)^2)+1/(x^2-1)=0

$\frac{4}{{(x + 1)}^{2}} - \frac{1}{{(x - 1)}^{2}} + \frac{1}{x^{2} - 1} = 0$

Factor each term.

Rewrite $1$ as $1^{2}$ .

$\frac{4}{{(x + 1)}^{2}} - \frac{1}{{(x - 1)}^{2}} + \frac{1}{x^{2} - 1^{2}} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{4}{{(x + 1)}^{2}} - \frac{1}{{(x - 1)}^{2}} + \frac{1}{(x + 1) (x - 1)} = 0$

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.

${(x + 1)}^{2}, {(x - 1)}^{2}, (x + 1) (x - 1), 1$

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 $x + 1$ are $(x + 1) \cdot (x + 1)$ , which is $x + 1$ multiplied by itself $2$ times.

$(x + 1) = (x + 1) \cdot (x + 1)$

$(x + 1)$ occurs $2$ times.

The factors for $x - 1$ are $(x - 1) \cdot (x - 1)$ , which is $x - 1$ multiplied by itself $2$ times.

$(x - 1) = (x - 1) \cdot (x - 1)$

$(x - 1)$ occurs $2$ times.

The factor for $x + 1$ is $x + 1$ itself.

$(x + 1) = x + 1$

$(x + 1)$ occurs $1$ time.

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

The LCM of ${(x + 1)}^{2}, {(x - 1)}^{2}, x + 1, x - 1$ is the result of multiplying all factors the greatest number of times they occur in either term.

${(x + 1)}^{2} {(x - 1)}^{2}$

Multiply each term in $\frac{4}{{(x + 1)}^{2}} - \frac{1}{{(x - 1)}^{2}} + \frac{1}{(x + 1) (x - 1)} = 0$ by ${(x + 1)}^{2} {(x - 1)}^{2}$ to eliminate the fractions.

Multiply each term in $\frac{4}{{(x + 1)}^{2}} - \frac{1}{{(x - 1)}^{2}} + \frac{1}{(x + 1) (x - 1)} = 0$ by ${(x + 1)}^{2} {(x - 1)}^{2}$ .

$\frac{4}{{(x + 1)}^{2}} ({(x + 1)}^{2} {(x - 1)}^{2}) - \frac{1}{{(x - 1)}^{2}} ({(x + 1)}^{2} {(x - 1)}^{2}) + \frac{1}{(x + 1) (x - 1)} ({(x + 1)}^{2} {(x - 1)}^{2}) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Simplify the left side.

Simplify each term.

Cancel the common factor of ${(x + 1)}^{2}$ .

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

$\frac{4}{{(x + 1)}^{2}} ({(x + 1)}^{2} ({(x - 1)}^{2})) - \frac{1}{{(x - 1)}^{2}} ({(x + 1)}^{2} {(x - 1)}^{2}) + \frac{1}{(x + 1) (x - 1)} ({(x + 1)}^{2} {(x - 1)}^{2}) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Cancel the common factor.

Rewrite the expression.

$4 {(x - 1)}^{2} - \frac{1}{{(x - 1)}^{2}} ({(x + 1)}^{2} {(x - 1)}^{2}) + \frac{1}{(x + 1) (x - 1)} ({(x + 1)}^{2} {(x - 1)}^{2}) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Cancel the common factor of ${(x - 1)}^{2}$ .

Move the leading negative in $- \frac{1}{{(x - 1)}^{2}}$ into the numerator.

$4 {(x - 1)}^{2} + \frac{- 1}{{(x - 1)}^{2}} ({(x + 1)}^{2} {(x - 1)}^{2}) + \frac{1}{(x + 1) (x - 1)} ({(x + 1)}^{2} {(x - 1)}^{2}) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

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

$4 {(x - 1)}^{2} + \frac{- 1}{{(x - 1)}^{2}} ({(x - 1)}^{2} {(x + 1)}^{2}) + \frac{1}{(x + 1) (x - 1)} ({(x + 1)}^{2} {(x - 1)}^{2}) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Cancel the common factor.

$4 {(x - 1)}^{2} + \frac{- 1}{{(x - 1)}^{2}} ({(x - 1)}^{2} {(x + 1)}^{2}) + \frac{1}{(x + 1) (x - 1)} ({(x + 1)}^{2} {(x - 1)}^{2}) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Rewrite the expression.

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + \frac{1}{(x + 1) (x - 1)} ({(x + 1)}^{2} {(x - 1)}^{2}) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Cancel the common factor of $(x + 1) (x - 1)$ .

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

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + \frac{1}{(x + 1) (x - 1)} ((x + 1) (x - 1) ((x + 1) (x - 1))) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Cancel the common factor.

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + \frac{1}{(x + 1) (x - 1)} ((x + 1) (x - 1) ((x + 1) (x - 1))) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Rewrite the expression.

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + (x + 1) (x - 1) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Expand $(x + 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + x (x - 1) + 1 (x - 1) = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 1$ to the left of $x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $1$ .

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + x^{2} - x + x - 1 = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Add $- x$ and $x$ .

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + x^{2} + 0 - 1 = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Add $x^{2}$ and $0$ .

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + x^{2} - 1 = 0 ({(x + 1)}^{2} {(x - 1)}^{2})$

Simplify the right side.

Multiply $0 ({(x + 1)}^{2} {(x - 1)}^{2})$ .

Multiply ${(x + 1)}^{2}$ by $0$ .

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + x^{2} - 1 = 0 {(x - 1)}^{2}$

Multiply $0$ by ${(x - 1)}^{2}$ .

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + x^{2} - 1 = 0$

Solve the equation.

Factor the left side of the equation.

Regroup terms.

$4 {(x - 1)}^{2} - {(x + 1)}^{2} + x^{2} - 1 = 0$

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

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

$4 {(x - 1)}^{2} - ({(x + 1)}^{2}) + x^{2} - 1 = 0$

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

$4 {(x - 1)}^{2} - ({(x + 1)}^{2}) - 1 (- x^{2}) - 1 = 0$

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

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

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

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

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

$4 {(x - 1)}^{2} - ({(x + 1)}^{2} - x^{2} + 1) = 0$

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

$4 {(x - 1)}^{2} - ((x + 1) (x + 1) - x^{2} + 1) = 0$

Expand $(x + 1) (x + 1)$ using the FOIL Method.

Apply the distributive property.

$4 {(x - 1)}^{2} - (x (x + 1) + 1 (x + 1) - x^{2} + 1) = 0$

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Multiply $x$ by $1$ .

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

Multiply $x$ by $1$ .

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

Multiply $1$ by $1$ .

$4 {(x - 1)}^{2} - (x^{2} + x + x + 1 - x^{2} + 1) = 0$

Add $x$ and $x$ .

$4 {(x - 1)}^{2} - (x^{2} + 2 x + 1 - x^{2} + 1) = 0$

Subtract $x^{2}$ from $x^{2}$ .

$4 {(x - 1)}^{2} - (2 x + 1 + 0 + 1) = 0$

Add $2 x$ and $0$ .

$4 {(x - 1)}^{2} - (2 x + 1 + 1) = 0$

Add $1$ and $1$ .

$4 {(x - 1)}^{2} - (2 x + 2) = 0$

Factor.

Factor $2$ out of $2 x + 2$ .

Factor $2$ out of $2 x$ .

$4 {(x - 1)}^{2} - (2 (x) + 2) = 0$

Factor $2$ out of $2$ .

$4 {(x - 1)}^{2} - (2 (x) + 2 (1)) = 0$

Factor $2$ out of $2 (x) + 2 (1)$ .

$4 {(x - 1)}^{2} - (2 (x + 1)) = 0$

Remove unnecessary parentheses.

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

Multiply $- 1$ by $2$ .

$4 {(x - 1)}^{2} - 2 (x + 1) = 0$

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

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

$2 (2 {(x - 1)}^{2}) - 2 (x + 1) = 0$

Factor $2$ out of $- 2 (x + 1)$ .

$2 (2 {(x - 1)}^{2}) + 2 (- (x + 1)) = 0$

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

$2 (2 {(x - 1)}^{2} - (x + 1)) = 0$

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$2 (2 ((x - 1) (x - 1)) - (x + 1)) = 0$

Expand $(x - 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$2 (2 (x (x - 1) - 1 (x - 1)) - (x + 1)) = 0$

Apply the distributive property.

$2 (2 (x \cdot x + x \cdot - 1 - 1 (x - 1)) - (x + 1)) = 0$

Apply the distributive property.

$2 (2 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) - (x + 1)) = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$2 (2 (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) - (x + 1)) = 0$

Move $- 1$ to the left of $x$ .

$2 (2 (x^{2} - 1 x - 1 x - 1 \cdot - 1) - (x + 1)) = 0$

Rewrite $- 1 x$ as $- x$ .

$2 (2 (x^{2} - x - 1 x - 1 \cdot - 1) - (x + 1)) = 0$

Rewrite $- 1 x$ as $- x$ .

$2 (2 (x^{2} - x - x - 1 \cdot - 1) - (x + 1)) = 0$

Multiply $- 1$ by $- 1$ .

$2 (2 (x^{2} - x - x + 1) - (x + 1)) = 0$

Subtract $x$ from $- x$ .

$2 (2 (x^{2} - 2 x + 1) - (x + 1)) = 0$

Apply the distributive property.

$2 (2 x^{2} + 2 (- 2 x) + 2 \cdot 1 - (x + 1)) = 0$

Simplify.

Multiply $- 2$ by $2$ .

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

Multiply $2$ by $1$ .

$2 (2 x^{2} - 4 x + 2 - (x + 1)) = 0$

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

$2 (2 x^{2} - 4 x + 2 - x - 1) = 0$

Subtract $x$ from $- 4 x$ .

$2 (2 x^{2} - 5 x + 2 - 1) = 0$

Subtract $1$ from $2$ .

$2 (2 x^{2} - 5 x + 1) = 0$

Divide each term in $2 (2 x^{2} - 5 x + 1) = 0$ by $2$ and simplify.

Divide each term in $2 (2 x^{2} - 5 x + 1) = 0$ by $2$ .

$\frac{2 (2 x^{2} - 5 x + 1)}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (2 x^{2} - 5 x + 1)}{2} = \frac{0}{2}$

Divide $2 x^{2} - 5 x + 1$ by $1$ .

$2 x^{2} - 5 x + 1 = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$2 x^{2} - 5 x + 1 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = - 5$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (2 \cdot 1)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 5$ to the power of $2$ .

$x = \frac{5 \pm \sqrt{25 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{5 \pm \sqrt{25 - 8 \cdot 1}}{2 \cdot 2}$

Multiply $- 8$ by $1$ .

$x = \frac{5 \pm \sqrt{25 - 8}}{2 \cdot 2}$

Subtract $8$ from $25$ .

$x = \frac{5 \pm \sqrt{17}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{5 \pm \sqrt{17}}{4}$

The final answer is the combination of both solutions.

$x = \frac{5 + \sqrt{17}}{4}, \frac{5 - \sqrt{17}}{4}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5 + \sqrt{17}}{4}, \frac{5 - \sqrt{17}}{4}$

Decimal Form:

$x = 2.28077640 \dots, 0.21922359 \dots$

Solve for x 4/((x+1)^2)-1/((x-1)^2)+1/(x^2-1)=0

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