# Solving Rational Equations (+ Questions)

The rational equations have the form , where P(x) and Q(x) are polynomials. The solution set of such an equation contains all the values of x which satisfy the equation.

We can solve a rational equation by transforming it to a simple equation using the least common denominator or substitution method. We may end up with more than one equation. We solve these and discard any extraneous solutions. We may optionally validate each solution by evaluating the original equation.

## Solution Steps

Let us see an example.

Solve

## Step 1: Find Equation Domain

The domain of a rational equation excludes all numbers for which any of the constituent rational functions is undefined. For a equation of the form , we exclude all values of x for which Q(x) = 0.

In our example, function is not defined for x = − 2So the domain of the equation is x ≠ − 2 or:

Domain = ℝ − { − 2}

### What is Domain?

A rational equation is composed of rational functions. There may be some real numbers where these functions are not defined. The domain of the equation excludes these numbers.

## Step 2: Simplify (LCD / Substitution)

### LCD Method

We simplify this equation using the LCD method. We find the least common denominator (LCD) and multiply it to both sides of the equation. Multiplication removes all the denominators in the equation.

The LCD of rational functions in the example equation is (x + 2).

We multiply both sides by it to simplify the equation:

x2 + x − 2 + 2x + 4 = 0

x2 + 3x + 2 = 0

## Step 3: Solve Simplified Equation

The solution involves factoring the polynomial to find the roots. Factoring this simple quadratic equation, we get:

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

x = − 1 and x = − 2 are solutions of this equation.

## Step 4: Remove Extraneous Solutions

While solving equations, we simplify them using transformations. However, the solution of the simplified equation may be outside the domain of the original equation. It is called an extraneous solution. We reject such solutions.

As you can see x = − 2 is outside the domain of original equation. We introduced this extraneous solution. Therefore we discard it as a solution to the original equation.

The only solution to our equation is x = { − 1}.

### ✩ Discard Extraneous Solutions (Important!)

• Transformations like multiplying with the least common denominator or substitution may introduce extraneous solutions.
• These appear to solve the equation but are outside its domain.
• Do not forget to exclude these from the solution set.

## Step 5: Verify Solutions (Optional)

Verify x = − 1.

Evaluating left-hand side (LHS) of the equation for x = − 1 we get:

LHS

= − 2 + 2 = 0 = RHS.

It is an acceptable solution.

Verify x = − 2.

Evaluating left-hand side (LHS) of the equation for x = − 2 we get:

LHS

Undefined.

x = − 2 is not a solution.

Let us see another example.

## Example – Substitution Method

It is easy to solve some rational equations by substitution. In this method, we replace some terms with a new variable. We then solve the transformed equation. The resulting solutions often result in multiple equations in the original variable. Let us solve one equation.

Solve

### Step 1. Find Equation Domain

What is the domain of this equation?

Domain excludes all x such that x3 − x2 = x2(x − 1) = 0.

So x = 0 and x = 1 are not in the domain of the equation.

Domain = ℝ − {0, 1}

### Step 2. Simplify (LCD / Substitution)

Let us substitute (x3 − x2) by y in the original equation:

y is the least common denominator here. Multiplying both sides by it, we get:

y2 − 8 = 2y

y2 − 2y − 8 = 0

### Step 3. Solve Simplified Equation

We easily factorize the quadratic equation as:

(y − 4)(y + 2) = 0

The solutions are y = 4, y = − 2These result in two equations:

1. x3 − x2 = 4
2. x3 − x2 = − 2.

Solving the first equation.

x3 − x2 = 4

x3 − x2 − 4 = 0

To factorize the cubic polynomial, we split −x2 term as −x2 = − 2x2 + x2:

x3 − 2x2 + x2 − 22 = 0

x2(x − 2) + (x − 2)(x + 2) = 0

(x − 2)(x2 + x + 2) = 0

The quadratic polynomial in the above equation cannot be factorized further (why?). So the only solution is x = 2.

Solving the second equation.

We write −x2 = − 2x2 + x2 and factorize.

x3 − x2 + 2 = 0

x3 − 2x2 + x2 + 2 = 0

x3 + x2 − 2x2 + 2 = 0

x2(x + 1) − 2(x2 − 1) = 0

x2(x + 1) − 2(x − 1)(x + 1) = 0

(x + 1)(x2 − 2x + 2) = 0

We cannot factor the quadratic polynomial above. The only solution is x = − 1.

We have two possible solutions to the original equation: x = 2, x = − 1

### Step 4: Remove Extraneous Solutions

Both the solutions are acceptable as they are in the domain of the equation.

### Step 5: Verify Solutions (Optional)

Let us verify x = − 1. We evaluate the left-hand side (LHS) of the equation:

LHS
= − 2 + 4 = 2 = RHS

The equation is balanced. x = − 1 is a solution.

Let us now verify x = 2.

Again evaluating the left-hand side, we get:

LHS

= 4 − 8/4 = 4 − 2 = 2 = RHS

x = 2 also satisfies the original equation.

Solution Set = { − 1, 2}

## Questions

### 1 Question

Find the domain of

For what values of x is undefined?

### 2 Question

Find the domain of

Are there any values of x for which the equation is undefined?

### 3 Question

Solve:

Can you factorize the quadratic denominator and solve the equation?

### 4 Question

Solve:

Is it possible to get a perfect square in the first bracket?

### 5 Question

Solve:

What is the domain and LCD for this equation? Can you multiply both sides by LCD?

The domain of this equation is same as that of the rational function . The remaining functions are defined for all the real numbers.

Now domain of f(x) excludes all x such that x3 − x2 = x2(x − 1) = 0 or x = 0 and x = 1.

Therefore domain of the equation ( and f(x) ) is:

Domain = ℝ − {0, 1}

Equation is sum of two rational functions. The function is not defined for x + 1 = 0 or x = − 1. Domain of f(x) is all real numbers except −1. We write it as:

Domainf(x) = ℝ − { − 1}

The other function is not defined for 2x − 1 = 0 or x = 1/2. Domain of g(x) is all real numbers except 1/2.

Domaing(x) = ℝ − {1/2}.

So the domain of equation is:

Domain = ℝ − { − 1, 1/2}

Second denominator is a quadratic polynomial. We factorize it by splitting the middle term.

2x2 + x − 1

= 2x(x + 1) − (x + 1) = (2x − 1)(x + 1)

This denominator is zero for x = − 1 and x = 1/2. So the domain of the equation is real numbers without these values.

Domain = ℝ − { − 1, 1/2}

Least common denominator (LCD) is (x + 1)(2x − 1). Multiplying both sides of the equation with it:

x(2x − 1) − 3 + 2(x + 1) = 0

2x2 − x − 3 + 2x + 2 = 0

2x2 + x − 1 = 0

(2x − 1)(x + 1) = 0

x = − 1 and x = 1/2 are solutions of the simplified equation.

The above solutions are not in the domain of the original equation. So we reject these solutions. There are no solutions in this case.

Solution Set = ϕ

Notice that the simplified equation is one of the denominators of the original equation. Any solution of this equation results in a zero denominator of the original equation.

1/x is not defined for x = 0. So the domain of equation = ℝ − {0}

The first bracket contains two square terms. These are squares of terms in the second bracket. This suggests that we should try to get a perfect square in the first bracket. To do so, we add and subtract 2 to the terms in the first bracket.

We get a simple quadratic equation after substituting by y.

2[y2 − 2] − 7y + 9 = 0

2y2 − 4 − 7y + 9 = 0

2y2 − 7y + 5 = 0

(y − 1)(2y − 5) = 0

The two solutions are: y = 1, y = 5/2.

Solving the first equation, y = 1:

x2 + 1 = x

x2 − x + 1 = 0

The discriminant of this quadratic is negative. This equation has no real solution.

Solving the other equation, y = 5/2:

Multiplying both sides by 2x:

2x2 + 2 = 5x

2x2 − 5x + 2 = 0

2x2 − 4x − x + 2 = 0

2x(x − 2) − (x − 2) = 0

(x − 2)(2x − 1) = 0

The solutions are x = 2, x = 1/2

As all the values are in the domain of the equation, they are valid. Let us verify them.

Putting x = 2 in the left-hand side (LHS) of the equation:

LHS

= 3 + ( − 6/2) = 3 − 3 = 0

x = 2 solves the equation.

Verify that x = 1/2 also solves the equation.

Solution Set = {1/2, 2}

The equation is not defined for x = 0.

Domain = ℝ − {0}

x2 is the least common denominator here. Multiplying both sides by it we get:

3x3 + 3 − 7x2 − 7x = 0

3x3 − 7x2 − 7x + 3 = 0

We need to factorize the above cubic polynomial. We split −7x2 as −7x2 = 3x2 − 10x2:

3x3 + 3x2 − 10x2 − 7x + 3 = 0

3x2(x + 1) − 10x2 − 7x + 3 = 0

Again splitting −7x = − 10x + 3x:

3x2(x + 1) − 10x2 − 10x + 3x + 3 = 0

3x2(x + 1) − 10x(x + 1) + 3(x + 1) = 0

(x + 1)(3x2 − 10x + 3) = 0

Finally splitting the middle term of the quadratic as −10x = − 9x − x:

(x + 1)(3x2 − 9x − x + 3) = 0

(x + 1)(3x(x − 3) − (x − 3)) = 0

(x + 1)(x − 3)(3x − 1) = 0

Possible solutions: x = − 1, x = 3, x = 1/3

All the above solutions are valid for the original equation as they are in its domain. They result in non-zero denominators. Let us verify x = 3. Evaluating the left hand side of the equation we get:

9 + (1/3) − 7 − (7/3)

2 − (6/3) = 2 − 2 = 0

Verify that other solutions are also valid.

Solution Set of original equation is { − 1, 1/3, 3}.

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