Solve (4)/((x)^(2)-9)-(x+1)/(x-3)=1 | 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

$$\frac{ 4 }{ { x }^{ 2 } -9 } - \frac{ x+1 }{ x-3 } =1$$

Solve for x (complex solution)

$x=\sqrt{6}-1\approx 1.449489743$
$x=-\left(\sqrt{6}+1\right)\approx -3.449489743$

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 $-3,3$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-3\right)\left(x+3\right)$, the least common multiple of $x^{2}-9,x-3$.

$$4-\left(x+3\right)\left(x+1\right)=\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply $x+3$ by $x+1$ and combine like terms.

$$4-\left(x^{2}+4x+3\right)=\left(x-3\right)\left(x+3\right)$$

To find the opposite of $x^{2}+4x+3$, find the opposite of each term.

$$4-x^{2}-4x-3=\left(x-3\right)\left(x+3\right)$$

Subtract $3$ from $4$ to get $1$.

$$1-x^{2}-4x=\left(x-3\right)\left(x+3\right)$$

Consider $\left(x-3\right)\left(x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.

$$1-x^{2}-4x=x^{2}-9$$

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

$$1-x^{2}-4x-x^{2}=-9$$

Combine $-x^{2}$ and $-x^{2}$ to get $-2x^{2}$.

$$1-2x^{2}-4x=-9$$

Add $9$ to both sides.

$$1-2x^{2}-4x+9=0$$

Add $1$ and $9$ to get $10$.

$$10-2x^{2}-4x=0$$

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.

$$-2x^{2}-4x+10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $-4$ for $b$, and $10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-\left(-4\right)±\sqrt{16+8\times 10}}{2\left(-2\right)}$$

Multiply $8$ times $10$.

$$x=\frac{-\left(-4\right)±\sqrt{16+80}}{2\left(-2\right)}$$

Add $16$ to $80$.

$$x=\frac{-\left(-4\right)±\sqrt{96}}{2\left(-2\right)}$$

Take the square root of $96$.

$$x=\frac{-\left(-4\right)±4\sqrt{6}}{2\left(-2\right)}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±4\sqrt{6}}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{4±4\sqrt{6}}{-4}$$

Now solve the equation $x=\frac{4±4\sqrt{6}}{-4}$ when $±$ is plus. Add $4$ to $4\sqrt{6}$.

$$x=\frac{4\sqrt{6}+4}{-4}$$

Divide $4+4\sqrt{6}$ by $-4$.

$$x=-\left(\sqrt{6}+1\right)$$

Now solve the equation $x=\frac{4±4\sqrt{6}}{-4}$ when $±$ is minus. Subtract $4\sqrt{6}$ from $4$.

$$x=\frac{4-4\sqrt{6}}{-4}$$

Divide $4-4\sqrt{6}$ by $-4$.

$$x=\sqrt{6}-1$$

The equation is now solved.

$$x=-\left(\sqrt{6}+1\right)$$ $$x=\sqrt{6}-1$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-3,3$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-3\right)\left(x+3\right)$, the least common multiple of $x^{2}-9,x-3$.

$$4-\left(x+3\right)\left(x+1\right)=\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply $x+3$ by $x+1$ and combine like terms.

$$4-\left(x^{2}+4x+3\right)=\left(x-3\right)\left(x+3\right)$$

To find the opposite of $x^{2}+4x+3$, find the opposite of each term.

$$4-x^{2}-4x-3=\left(x-3\right)\left(x+3\right)$$

Subtract $3$ from $4$ to get $1$.

$$1-x^{2}-4x=\left(x-3\right)\left(x+3\right)$$

Consider $\left(x-3\right)\left(x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.

$$1-x^{2}-4x=x^{2}-9$$

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

$$1-x^{2}-4x-x^{2}=-9$$

Combine $-x^{2}$ and $-x^{2}$ to get $-2x^{2}$.

$$1-2x^{2}-4x=-9$$

Subtract $1$ from both sides.

$$-2x^{2}-4x=-9-1$$

Subtract $1$ from $-9$ to get $-10$.

$$-2x^{2}-4x=-10$$

Divide both sides by $-2$.

$$\frac{-2x^{2}-4x}{-2}=-\frac{10}{-2}$$

Dividing by $-2$ undoes the multiplication by $-2$.

$$x^{2}+\left(-\frac{4}{-2}\right)x=-\frac{10}{-2}$$

Divide $-4$ by $-2$.

$$x^{2}+2x=-\frac{10}{-2}$$

Divide $-10$ by $-2$.

$$x^{2}+2x=5$$

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.

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

Square $1$.

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

Add $5$ to $1$.

$$x^{2}+2x+1=6$$

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

$$\left(x+1\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{6}$$

Simplify.

$$x+1=\sqrt{6}$$ $$x+1=-\sqrt{6}$$

Subtract $1$ from both sides of the equation.

$$x=\sqrt{6}-1$$ $$x=-\sqrt{6}-1$$

Solve for x

$x=\sqrt{6}-1\approx 1.449489743$
$x=-\sqrt{6}-1\approx -3.449489743$

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 $-3,3$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-3\right)\left(x+3\right)$, the least common multiple of $x^{2}-9,x-3$.

$$4-\left(x+3\right)\left(x+1\right)=\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply $x+3$ by $x+1$ and combine like terms.

$$4-\left(x^{2}+4x+3\right)=\left(x-3\right)\left(x+3\right)$$

To find the opposite of $x^{2}+4x+3$, find the opposite of each term.

$$4-x^{2}-4x-3=\left(x-3\right)\left(x+3\right)$$

Subtract $3$ from $4$ to get $1$.

$$1-x^{2}-4x=\left(x-3\right)\left(x+3\right)$$

Consider $\left(x-3\right)\left(x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.

$$1-x^{2}-4x=x^{2}-9$$

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

$$1-x^{2}-4x-x^{2}=-9$$

Combine $-x^{2}$ and $-x^{2}$ to get $-2x^{2}$.

$$1-2x^{2}-4x=-9$$

Add $9$ to both sides.

$$1-2x^{2}-4x+9=0$$

Add $1$ and $9$ to get $10$.

$$10-2x^{2}-4x=0$$

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.

$$-2x^{2}-4x+10=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $-4$ for $b$, and $10$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-\left(-4\right)±\sqrt{16+8\times 10}}{2\left(-2\right)}$$

Multiply $8$ times $10$.

$$x=\frac{-\left(-4\right)±\sqrt{16+80}}{2\left(-2\right)}$$

Add $16$ to $80$.

$$x=\frac{-\left(-4\right)±\sqrt{96}}{2\left(-2\right)}$$

Take the square root of $96$.

$$x=\frac{-\left(-4\right)±4\sqrt{6}}{2\left(-2\right)}$$

The opposite of $-4$ is $4$.

$$x=\frac{4±4\sqrt{6}}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{4±4\sqrt{6}}{-4}$$

Now solve the equation $x=\frac{4±4\sqrt{6}}{-4}$ when $±$ is plus. Add $4$ to $4\sqrt{6}$.

$$x=\frac{4\sqrt{6}+4}{-4}$$

Divide $4+4\sqrt{6}$ by $-4$.

$$x=-\left(\sqrt{6}+1\right)$$

Now solve the equation $x=\frac{4±4\sqrt{6}}{-4}$ when $±$ is minus. Subtract $4\sqrt{6}$ from $4$.

$$x=\frac{4-4\sqrt{6}}{-4}$$

Divide $4-4\sqrt{6}$ by $-4$.

$$x=\sqrt{6}-1$$

The equation is now solved.

$$x=-\left(\sqrt{6}+1\right)$$ $$x=\sqrt{6}-1$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-3,3$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-3\right)\left(x+3\right)$, the least common multiple of $x^{2}-9,x-3$.

$$4-\left(x+3\right)\left(x+1\right)=\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply $x+3$ by $x+1$ and combine like terms.

$$4-\left(x^{2}+4x+3\right)=\left(x-3\right)\left(x+3\right)$$

To find the opposite of $x^{2}+4x+3$, find the opposite of each term.

$$4-x^{2}-4x-3=\left(x-3\right)\left(x+3\right)$$

Subtract $3$ from $4$ to get $1$.

$$1-x^{2}-4x=\left(x-3\right)\left(x+3\right)$$

Consider $\left(x-3\right)\left(x+3\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $3$.

$$1-x^{2}-4x=x^{2}-9$$

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

$$1-x^{2}-4x-x^{2}=-9$$

Combine $-x^{2}$ and $-x^{2}$ to get $-2x^{2}$.

$$1-2x^{2}-4x=-9$$

Subtract $1$ from both sides.

$$-2x^{2}-4x=-9-1$$

Subtract $1$ from $-9$ to get $-10$.

$$-2x^{2}-4x=-10$$

Divide both sides by $-2$.

$$\frac{-2x^{2}-4x}{-2}=-\frac{10}{-2}$$

Dividing by $-2$ undoes the multiplication by $-2$.

$$x^{2}+\left(-\frac{4}{-2}\right)x=-\frac{10}{-2}$$

Divide $-4$ by $-2$.

$$x^{2}+2x=-\frac{10}{-2}$$

Divide $-10$ by $-2$.

$$x^{2}+2x=5$$

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.

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

Square $1$.

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

Add $5$ to $1$.

$$x^{2}+2x+1=6$$

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

$$\left(x+1\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{6}$$

Simplify.

$$x+1=\sqrt{6}$$ $$x+1=-\sqrt{6}$$

Subtract $1$ from both sides of the equation.

$$x=\sqrt{6}-1$$ $$x=-\sqrt{6}-1$$