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Extraneous Solutions

By Debi DalPezzo

When solving all Rational Equations there is a possibility that one of the solutions will produce an untrue statement when substituted back into the problem. These solutions that don’t work are called Extraneous Solutions.

Why is this concept useful?

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that don't satisfy the original form of the equation because they produce untrue statements or are excluded values that make a denominator equal to 0.

Where does this concept fit into the curriculum?

Algebra 2

How can we use the concept:

We should always check our solutions upon solving rational equations to make sure we don't have any Extraneous Solutions.

Sample Math Problems

1. Find the extraneous solution of the following equation.

2x2−1−1x−1=1\frac{2}{x^{2} -1} -\frac{1}{x-1} =1

Answer:

Step 1: Find the LCD to clear the fraction:

(x2−1)∗(2x2−1−1x−1=1)=2−x−1=x2−1(x^2 - 1)*(\frac{2}{x^2 - 1} - \frac{1}{x-1} = 1) = 2 - x - 1 = x^2 -1

Step 2: Combine Like Terms:

−x+1=x2−1-x+1=x^2-1

Step 3: Set everything equal to zero:

0=x2+x−20 = x^2 + x - 2

Step 4: Factor if possible or use the Quadratic Formula to find the solutions to this equation:

x2+x−2=0x^2 + x - 2 = 0
(x−1)(x−2)(x-1)(x-2)

Therefore the solutions are 1 and -2, but what do you notice about the solutions. Can 1 be a solution? If you put 1 back into the original problem your denominator will be zero, so the solution {1} is an extraneous solution!

2. Find the extraneous solution of the following equation.

1 = x3+6x23x21\ =\ \frac{x^{3} +6x^{2}}{3x^{2}}

Answer:

Step 1: Find the LCD:  3x2 = x3+ 6x2Step 2: CLT: None: Step 3: Set equal to zero:  0 = x3+ 3x2Step 4: Factor:  x2(x + 3) = 0 \begin{array}{l} Step\ 1:\ Find\ the\ LCD:\ \ 3x^{2} \ =\ x^{3} +\ 6x^{2}\\ Step\ 2:\ CLT:\ None:\ \\ Step\ 3:\ Set\ equal\ to\ zero:\ \ 0\ =\ x^{3} +\ 3x^{2}\\ Step\ 4:\ Factor:\ \ x^{2}( x\ +\ 3) \ =\ 0 \end{array}

Therefore x = 0 and x = -3. Check your solutions. Do they both work in the original equation?

No, x = 0 is an extraneous solution!


3. Find the extraneous solution of the following equation.

xx+5−2x−9=−11x+15x2−4x−45\frac{x}{x+5} -\frac{2}{x-9} =\frac{-11x+15}{x^{2} -4x-45}

Answer:

Step 1: Find the LCD:  x2−9x−2x−10= −11x+15Step 2: CLT: x2−11x−10= −11x+15Step 3: Set equal to zero:  x2− 25= 0Step 4: Factor:  (x + 5)(x − 5)= 0 \begin{array}{l} Step\ 1:\ Find\ the\ LCD:\ \ x^{2} -9x-2x-10=\ -11x+15\\ Step\ 2:\ CLT:\ x^{2} -11x-10=\ -11x+15\\ Step\ 3:\ Set\ equal\ to\ zero:\ \ x^{2} -\ 25=\ 0\\ Step\ 4:\ Factor:\ \ ( x\ +\ 5)( x\ -\ 5) =\ 0 \end{array}

Which of these factors are extraneous? Correct {-5}.

4. Find the extraneous solution of the following equation.

yy−8−2y−4=−3y+56y2−12y+32\frac{y}{y-8} -\frac{2}{y-4} =\frac{-3y+56}{y^{2} -12y+32}

Answer

Step 1: Find the LCD:  y2−4y−2y+16= −3y+56Step 2: CLT: y2−6y+16= −3y+56Step 3: Set equal to zero:  y2− 3y−40= 0Step 4: Factor:  (y − 8)(y + 5)= 0\begin{array}{l} Step\ 1:\ Find\ the\ LCD:\ \ y^{2} - 4y -2y + 16 =\ -3y+56\\ Step\ 2:\ CLT:\ y^{2} -6y + 16 =\ -3y +56\\ Step\ 3:\ Set\ equal\ to\ zero:\ \ y^{2} -\ 3y -40 =\ 0\\ Step\ 4:\ Factor:\ \ ( y\ -\ 8)( y\ +\ 5) =\ 0 \end{array}

Which of these factors are extraneous?  Correct {8}.

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Practice Math Problems

1. Find the extraneous solution(s) of the following equation.

yy−2+2y−4=4y−12y2−6y+8\frac{y}{y-2} +\frac{2}{y-4} =\frac{4y-12}{y^{2} -6y+8}

2. Find the extraneous solution of the following equation.

5x+3=xx+2+2x2+5x+6\frac{5}{x+3} =\frac{x}{x+2} +\frac{2}{x^{2} +5x+6}

3. Find the extraneous solution of the following equation.

8x2x2−9−4xx+3=6x−3\frac{8x^{2}}{x^{2} -9} -\frac{4x}{x+3} =\frac{6}{x-3}

4. Find the extraneous solution of the following equation.

5yy+1=4−5y+1\frac{5y}{y+1} =4-\frac{5}{y+1}

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