Solve Rational Equations And Check For Extraneous Solutions

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When dealing with equations involving rational expressions, it's important to solve for the variable and also check for extraneous solutions. Extraneous solutions are values that arise during the solving process that appear to be solutions, but when substituted back into the original equation, they make the equation undefined, often by causing division by zero. In this comprehensive guide, we will walk through the process of solving the given rational equation step-by-step, emphasizing the critical steps to identify and eliminate any extraneous solutions. Our aim is to provide a clear, in-depth understanding of how to handle such equations, ensuring accuracy and confidence in your mathematical skills. This detailed approach will not only help you solve the specific equation at hand but also equip you with the knowledge to tackle similar problems effectively.

Step-by-Step Solution

Let's consider the equation:

x+5x=2x+13+26x2+13x\frac{x+5}{x} = \frac{2}{x+13} + \frac{26}{x^2+13x}

The first step in solving this equation is to eliminate the fractions. To do this, we need to find the least common denominator (LCD) of the denominators involved. In this case, the denominators are x, x + 13, and x2 + 13x. Notice that x2 + 13x can be factored as x(x + 13). Thus, the LCD is x(x + 13). Identifying the LCD is a crucial step as it allows us to clear the fractions, making the equation easier to solve. A clear understanding of this process not only simplifies the algebraic manipulation but also sets the stage for correctly identifying any potential extraneous solutions later on.

Multiply both sides of the equation by the LCD, x(x + 13), to eliminate the denominators:

x(x+13)â‹…x+5x=x(x+13)â‹…(2x+13+26x(x+13))x(x+13) \cdot \frac{x+5}{x} = x(x+13) \cdot \left( \frac{2}{x+13} + \frac{26}{x(x+13)} \right)

This step involves distributing the LCD across all terms in the equation. When we multiply each term by x(x + 13), we carefully cancel out the common factors in the denominators. This process effectively clears the fractions, transforming the equation into a more manageable form, typically a polynomial equation. It is essential to ensure that the LCD is multiplied correctly with each term to maintain the equation's balance and accuracy. This careful manipulation is crucial for arriving at the correct solution while avoiding common algebraic errors.

Simplifying, we get:

(x+13)(x+5)=2x+26(x+13)(x+5) = 2x + 26

Now, we expand the left side of the equation by multiplying the two binomials. This involves applying the distributive property (often remembered by the acronym FOIL: First, Outer, Inner, Last) to each term. The expansion of (x + 13)(x + 5) results in a quadratic expression. Expanding the product accurately is a critical step because any error here can lead to an incorrect quadratic equation and, consequently, the wrong solutions. This process transforms the equation into a standard polynomial form, which is easier to manipulate and solve.

Expanding the left side, we have:

x2+18x+65=2x+26x^2 + 18x + 65 = 2x + 26

Next, we rearrange the equation to set it equal to zero. This step is fundamental in solving quadratic equations because it allows us to use methods such as factoring, completing the square, or the quadratic formula. To set the equation to zero, we subtract the terms on the right side from both sides, effectively moving all terms to one side. This rearrangement not only prepares the equation for solving but also ensures that we have a standard form of a quadratic equation, which is necessary for applying the standard solution techniques. The accuracy of this step is vital for obtaining the correct solutions.

Subtracting 2x + 26 from both sides, we get:

x2+16x+39=0x^2 + 16x + 39 = 0

Now, we factor the quadratic equation. Factoring is a method of breaking down the quadratic expression into two binomial expressions that, when multiplied together, give the original quadratic. The goal is to find two numbers that add up to the coefficient of the x term (in this case, 16) and multiply to the constant term (in this case, 39). Factoring, if possible, is often the quickest way to solve a quadratic equation. However, it's important to note that not all quadratic equations can be factored easily, and sometimes other methods like the quadratic formula may be necessary. The ability to factor efficiently is a valuable skill in algebra.

Factoring the quadratic, we look for two numbers that multiply to 39 and add to 16. These numbers are 3 and 13. So, we can factor the quadratic as:

(x+3)(x+13)=0(x+3)(x+13) = 0

Using the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero, we set each factor equal to zero and solve for x. This property is a fundamental concept in algebra and is crucial for solving equations that are factored. By setting each factor to zero, we generate potential solutions for the equation. It is important to remember that these solutions are potential and must be checked for extraneous solutions later.

Setting each factor to zero gives us:

x+3=0orx+13=0x + 3 = 0 \quad \text{or} \quad x + 13 = 0

Solving these equations, we find the potential solutions for x. Each linear equation is solved by isolating x on one side. These values are potential solutions because they satisfy the factored equation, but it's essential to verify whether they also satisfy the original equation, especially when dealing with rational expressions. This verification step is crucial to identify and eliminate any extraneous solutions that may have been introduced during the solving process.

Solving for x in each equation, we get:

x=−3orx=−13x = -3 \quad \text{or} \quad x = -13

Checking for Extraneous Solutions

Now, it's crucial to check these potential solutions in the original equation to ensure they are valid and not extraneous. Extraneous solutions can arise when we perform operations like squaring both sides or, as in this case, multiplying by an expression that contains the variable. These operations can introduce solutions that do not satisfy the original equation. Therefore, checking each potential solution in the original equation is a necessary step to ensure the accuracy of our results.

Recall the original equation:

x+5x=2x+13+26x2+13x\frac{x+5}{x} = \frac{2}{x+13} + \frac{26}{x^2+13x}

First, let's check x = -3. We substitute -3 for x in the original equation and simplify both sides. This substitution must be done carefully to ensure accuracy, paying close attention to signs and order of operations. The goal is to determine whether the left-hand side (LHS) of the equation equals the right-hand side (RHS) when x = -3. If the LHS and RHS are equal, then -3 is a valid solution; otherwise, it is an extraneous solution.

Substituting x = -3:

−3+5−3=2−3+13+26(−3)2+13(−3)\frac{-3+5}{-3} = \frac{2}{-3+13} + \frac{26}{(-3)^2+13(-3)}

2−3=210+269−39\frac{2}{-3} = \frac{2}{10} + \frac{26}{9-39}

−23=15+26−30-\frac{2}{3} = \frac{1}{5} + \frac{26}{-30}

−23=15−1315-\frac{2}{3} = \frac{1}{5} - \frac{13}{15}

−23=315−1315-\frac{2}{3} = \frac{3}{15} - \frac{13}{15}

−23=−1015-\frac{2}{3} = \frac{-10}{15}

−23=−23-\frac{2}{3} = -\frac{2}{3}

Since the equation holds true, x = -3 is a valid solution. The fact that both sides of the equation are equal confirms that -3 is indeed a solution to the original equation. This verification process is crucial to ensure that we do not include extraneous solutions in our final answer. Now, we proceed to check the other potential solution.

Now, let's check x = -13. We substitute -13 for x in the original equation. This step is similar to the previous one, but it's crucial to perform it carefully, especially when dealing with negative numbers. Substituting -13 involves replacing every instance of x in the original equation with -13 and then simplifying both sides to check for equality. If the substitution results in any undefined terms (such as division by zero), or if the left-hand side does not equal the right-hand side, then -13 is an extraneous solution.

Substituting x = -13:

−13+5−13=2−13+13+26(−13)2+13(−13)\frac{-13+5}{-13} = \frac{2}{-13+13} + \frac{26}{(-13)^2+13(-13)}

−8−13=20+26169−169\frac{-8}{-13} = \frac{2}{0} + \frac{26}{169-169}

The denominators on the right-hand side become zero, making the terms undefined. This immediately indicates that x = -13 is an extraneous solution because division by zero is undefined in mathematics. An extraneous solution is a value that appears to be a solution algebraically but does not satisfy the original equation. In this case, -13 cannot be a solution because it makes the equation undefined. This is a clear demonstration of why checking potential solutions in the original equation is an essential step in solving rational equations.

Thus, x = -13 is an extraneous solution.

Final Answer

Therefore, the only valid solution to the equation is:

x=−3x = -3

In summary, solving rational equations requires a methodical approach that includes finding the least common denominator, clearing fractions, solving the resulting equation, and, most importantly, checking for extraneous solutions. Extraneous solutions are values that arise during the solving process but do not satisfy the original equation, often due to division by zero. By meticulously checking each potential solution, we ensure that our final answer is accurate and valid. This comprehensive process not only solves the equation at hand but also reinforces the critical algebraic skills necessary for tackling more complex problems. Remember, accuracy and verification are key to success in mathematics.