Exercise 6 Page 695

We are asked to write a rational equation that has one solution and one extraneous solution. Let's start by recalling a few important definitions.

• Rational Equation: An equation that contains at least one rational expression.
• Rational Expression: A fraction whose numerator and denominator are polynomials.
• Extraneous Solution: A solution that is derived from the original equation, but it is not actually a solution.

When we end up with an extraneous solution in a rational equation, it is usually because one of our solutions is an excluded value. An excluded value is any number that would make the denominator of any fraction in the original equation equal to 0.

Rational Equation	Excluded Values
2/x+5=7	x=- 5
2x+3/x^2-4=x/x^2-4	x=- 2 and x=2
x-1/x^2-2x+1=4/x	x=0 and x=1

Writing an Equation

Keeping this in mind, we will start writing our equation by deciding what our excluded value will be and writing an expression for the denominator of at least one of our fractions. Excluded Value:& x= 1 Expression:& x- 1For simplicity, let's write a rational equation with two fractions that have the same denominator — the above expression. ?/x- 1=?/x- 1 Because we want our rational equation to have two solutions — one valid solution and one extraneous solution — we should obtain a quadratic equation after clearing the denominators. We can chose the first numerator arbitrarily. Then, the solutions of our equation will depend on the remaining numerator. x^2-3x/x-1=?/x-1 We can guarantee that x= 1 is a solution to our equation by ensuring that, when the denominators are cleared, the numerators will result in a quadratic equation with this value as one of its solutions. x^2-3x/x-1=?/x-1 ⇒ x^2-3x=? Let's substitute x= 1 into x^2-3x and evaluate. This will help us decide what expression to put on the right-hand side of our quadratic equation.

x^2-3x

Substitute

x= 1

1^2-3( 1)

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Evaluate

BaseOne

1^a=1

1-3(1)

MultByOne

a * 1=a

1-3

SubTerm

Subtract term

- 2

For x= 1 to be the solution of our quadratic equation, the right-hand side should also equal - 2 after substituting this value. There are many expressions that we could put on the right-hand side to meet this condition. Here are a few examples.

Quadratic Equations That Have x=1 as a Solution
x^2-3x=- 2
x^2-3x=x-3
x^2-3x=- x^2-x

Let's use the first of the above equations. The right-hand side of this equation will be the remaining numerator. x^2-3x/x-1=- 2/x-1 Now we know that x=1 is an extraneous solution of the equation. Also, because of the method that we used to choose the quadratic equation in the numerator, we know that there will be two unique solutions — one of which is the extraneous x=1.

Solving Our Rational Equation

To make sure that we have, in fact, written a rational equation that has one solution and one extraneous solution, we will solve our equation.

x^2-3x/x-1=- 2/x-1

MultEqn

LHS * (x-1)=RHS* (x-1)

x^2-3x=- 2

AddEqn

LHS+2=RHS+2

x^2-3x+2=0

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Factor

WriteDiff

Write as a difference

x^2-x-2x+2=0

FactorOut

Factor out x

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

FactorOut

Factor out - 2

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

FactorOut

Factor out (x-1)

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

ZeroProdProp

Use the Zero Product Property

lcx-1=0 & (I) x-2=0 & (II)

AddEqn

(I): LHS+1=RHS+1

lx=1 x-2=0

AddEqn

(II): LHS+2=RHS+2

lx=1 x=2

Now we will check our solutions by substituting them into the original equation. Let's start with x=1.

x^2-3x/x-1=- 2/x-1

Substitute

x= 1

1^2-3( 1)/1-1? =- 2/1-1

SubTerms

Subtract terms

1^2-3(1)/0? =- 2/0 *

We can see that the denominators of fractions are equal to 0 after substituting x=1, so it is an extraneous solution. Next, let's check x=2.

x^2-3x/x-1=- 2/x-1

Substitute

x= 2

2^2-3( 2)/2-1? =- 2/2-1

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Simplify

SubTerms

Subtract terms

2^2-3(2)/1? =- 2/1

DivByOne

a/1=a

2^2-3(2)? =- 2

CalcPowProd

Calculate power and product

4-6? =- 2

SubTerm

Subtract term

- 2=- 2 ✓

This solution is not an extraneous solution. We have successfully written an example of the requested rational equation.