Solve ((x)^(2)+3)/((x)^(2)-x-6)=(1)/(3-x) | 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

$$\frac{ { x }^{ 2 } +3 }{ { x }^{ 2 } -x-6 } = \frac{ 1 }{ 3-x }$$

Solve for x (complex solution)

$x=\frac{-\sqrt{19}i-1}{2}\approx -0.5-2.179449472i$
$x=\frac{-1+\sqrt{19}i}{2}\approx -0.5+2.179449472i$

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 $-2,3$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-3\right)\left(x+2\right)$, the least common multiple of $x^{2}-x-6,3-x$.

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

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

$$x^{2}+3=-2-x$$

Subtract $-2$ from both sides.

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

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

$$x^{2}+3+2=-x$$

Add $x$ to both sides.

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

Add $3$ and $2$ to get $5$.

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

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.

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $1$ for $b$, and $5$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-1±\sqrt{1^{2}-4\times 5}}{2}$$

Square $1$.

$$x=\frac{-1±\sqrt{1-4\times 5}}{2}$$

Multiply $-4$ times $5$.

$$x=\frac{-1±\sqrt{1-20}}{2}$$

Add $1$ to $-20$.

$$x=\frac{-1±\sqrt{-19}}{2}$$

Take the square root of $-19$.

$$x=\frac{-1±\sqrt{19}i}{2}$$

Now solve the equation $x=\frac{-1±\sqrt{19}i}{2}$ when $±$ is plus. Add $-1$ to $i\sqrt{19}$.

$$x=\frac{-1+\sqrt{19}i}{2}$$

Now solve the equation $x=\frac{-1±\sqrt{19}i}{2}$ when $±$ is minus. Subtract $i\sqrt{19}$ from $-1$.

$$x=\frac{-\sqrt{19}i-1}{2}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{19}i}{2}$$ $$x=\frac{-\sqrt{19}i-1}{2}$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-2,3$ since division by zero is not defined. Multiply both sides of the equation by $\left(x-3\right)\left(x+2\right)$, the least common multiple of $x^{2}-x-6,3-x$.

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

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

$$x^{2}+3=-2-x$$

Add $x$ to both sides.

$$x^{2}+3+x=-2$$

Subtract $3$ from both sides.

$$x^{2}+x=-2-3$$

Subtract $3$ from $-2$ to get $-5$.

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

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

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

Square $\frac{1}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=-5+\frac{1}{4}$$

Add $-5$ to $\frac{1}{4}$.

$$x^{2}+x+\frac{1}{4}=-\frac{19}{4}$$

Factor $x^{2}+x+\frac{1}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{19}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{19}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{19}i}{2}$$ $$x+\frac{1}{2}=-\frac{\sqrt{19}i}{2}$$

Subtract $\frac{1}{2}$ from both sides of the equation.

$$x=\frac{-1+\sqrt{19}i}{2}$$ $$x=\frac{-\sqrt{19}i-1}{2}$$