Solve 1/(x ^ 2 - 1) - 1/(x - 1) = 1/2 | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

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

Solve for x (complex solution)

$x=\sqrt{2}-1\approx 0.414213562$
$x=-\left(\sqrt{2}+1\right)\approx -2.414213562$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x-1,2$.

$$2-\left(2x+2\right)=\left(x-1\right)\left(x+1\right)$$

To find the opposite of $2x+2$, find the opposite of each term.

$$2-2x-2=\left(x-1\right)\left(x+1\right)$$

Subtract $2$ from $2$ to get $0$.

$$-2x=\left(x-1\right)\left(x+1\right)$$

Consider $\left(x-1\right)\left(x+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

$$-2x=x^{2}-1$$

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

$$-2x-x^{2}=-1$$

Add $1$ to both sides.

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

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-2$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Square $-2$.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-2\right)±\sqrt{4+4}}{2\left(-1\right)}$$

Add $4$ to $4$.

$$x=\frac{-\left(-2\right)±\sqrt{8}}{2\left(-1\right)}$$

Take the square root of $8$.

$$x=\frac{-\left(-2\right)±2\sqrt{2}}{2\left(-1\right)}$$

The opposite of $-2$ is $2$.

$$x=\frac{2±2\sqrt{2}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{2±2\sqrt{2}}{-2}$$

Now solve the equation $x=\frac{2±2\sqrt{2}}{-2}$ when $±$ is plus. Add $2$ to $2\sqrt{2}$.

$$x=\frac{2\sqrt{2}+2}{-2}$$

Divide $2+2\sqrt{2}$ by $-2$.

$$x=-\left(\sqrt{2}+1\right)$$

Now solve the equation $x=\frac{2±2\sqrt{2}}{-2}$ when $±$ is minus. Subtract $2\sqrt{2}$ from $2$.

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

Divide $2-2\sqrt{2}$ by $-2$.

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

The equation is now solved.

$$x=-\left(\sqrt{2}+1\right)$$ $$x=\sqrt{2}-1$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x-1,2$.

$$2-\left(2x+2\right)=\left(x-1\right)\left(x+1\right)$$

To find the opposite of $2x+2$, find the opposite of each term.

$$2-2x-2=\left(x-1\right)\left(x+1\right)$$

Subtract $2$ from $2$ to get $0$.

$$-2x=\left(x-1\right)\left(x+1\right)$$

Consider $\left(x-1\right)\left(x+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

$$-2x=x^{2}-1$$

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

$$-2x-x^{2}=-1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-x^{2}-2x=-1$$

Divide both sides by $-1$.

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

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\left(-\frac{2}{-1}\right)x=-\frac{1}{-1}$$

Divide $-2$ by $-1$.

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

Divide $-1$ by $-1$.

$$x^{2}+2x=1$$

Divide $2$, the coefficient of the $x$ term, by $2$ to get $1$. Then add the square of $1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=1+1^{2}$$

Square $1$.

$$x^{2}+2x+1=1+1$$

Add $1$ to $1$.

$$x^{2}+2x+1=2$$

Factor $x^{2}+2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+1\right)^{2}=2$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{2}$$

Simplify.

$$x+1=\sqrt{2}$$ $$x+1=-\sqrt{2}$$

Subtract $1$ from both sides of the equation.

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

Solve for x

$x=\sqrt{2}-1\approx 0.414213562$
$x=-\sqrt{2}-1\approx -2.414213562$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x-1,2$.

$$2-\left(2x+2\right)=\left(x-1\right)\left(x+1\right)$$

To find the opposite of $2x+2$, find the opposite of each term.

$$2-2x-2=\left(x-1\right)\left(x+1\right)$$

Subtract $2$ from $2$ to get $0$.

$$-2x=\left(x-1\right)\left(x+1\right)$$

Consider $\left(x-1\right)\left(x+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

$$-2x=x^{2}-1$$

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

$$-2x-x^{2}=-1$$

Add $1$ to both sides.

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

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $-2$ for $b$, and $1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)}}{2\left(-1\right)}$$

Square $-2$.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-\left(-2\right)±\sqrt{4+4}}{2\left(-1\right)}$$

Add $4$ to $4$.

$$x=\frac{-\left(-2\right)±\sqrt{8}}{2\left(-1\right)}$$

Take the square root of $8$.

$$x=\frac{-\left(-2\right)±2\sqrt{2}}{2\left(-1\right)}$$

The opposite of $-2$ is $2$.

$$x=\frac{2±2\sqrt{2}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{2±2\sqrt{2}}{-2}$$

Now solve the equation $x=\frac{2±2\sqrt{2}}{-2}$ when $±$ is plus. Add $2$ to $2\sqrt{2}$.

$$x=\frac{2\sqrt{2}+2}{-2}$$

Divide $2+2\sqrt{2}$ by $-2$.

$$x=-\left(\sqrt{2}+1\right)$$

Now solve the equation $x=\frac{2±2\sqrt{2}}{-2}$ when $±$ is minus. Subtract $2\sqrt{2}$ from $2$.

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

Divide $2-2\sqrt{2}$ by $-2$.

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

The equation is now solved.

$$x=-\left(\sqrt{2}+1\right)$$ $$x=\sqrt{2}-1$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-1,1$ since division by zero is not defined. Multiply both sides of the equation by $2\left(x-1\right)\left(x+1\right)$, the least common multiple of $x^{2}-1,x-1,2$.

$$2-\left(2x+2\right)=\left(x-1\right)\left(x+1\right)$$

To find the opposite of $2x+2$, find the opposite of each term.

$$2-2x-2=\left(x-1\right)\left(x+1\right)$$

Subtract $2$ from $2$ to get $0$.

$$-2x=\left(x-1\right)\left(x+1\right)$$

Consider $\left(x-1\right)\left(x+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

$$-2x=x^{2}-1$$

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

$$-2x-x^{2}=-1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-x^{2}-2x=-1$$

Divide both sides by $-1$.

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

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\left(-\frac{2}{-1}\right)x=-\frac{1}{-1}$$

Divide $-2$ by $-1$.

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

Divide $-1$ by $-1$.

$$x^{2}+2x=1$$

Divide $2$, the coefficient of the $x$ term, by $2$ to get $1$. Then add the square of $1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=1+1^{2}$$

Square $1$.

$$x^{2}+2x+1=1+1$$

Add $1$ to $1$.

$$x^{2}+2x+1=2$$

Factor $x^{2}+2x+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+1\right)^{2}=2$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{2}$$

Simplify.

$$x+1=\sqrt{2}$$ $$x+1=-\sqrt{2}$$

Subtract $1$ from both sides of the equation.

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