Solving Rational Equations Check For Extraneous Solutions

by Admin 58 views

In mathematics, solving equations is a fundamental skill. This article delves into the process of solving a specific type of equation: a rational equation. We will dissect the equation d2d−6−3d2−6d+9=d−23d−9{\frac{d}{2 d-6}-\frac{3}{d^2-6 d+9}=\frac{d-2}{3 d-9}}, providing a step-by-step solution and emphasizing the crucial step of verifying solutions to avoid extraneous roots. This guide is designed to enhance your understanding of algebraic manipulations and critical thinking in problem-solving.

Understanding Rational Equations

Rational equations, at their core, involve algebraic fractions. These fractions contain polynomials in both the numerator and the denominator. The equation d2d−6−3d2−6d+9=d−23d−9{\frac{d}{2 d-6}-\frac{3}{d^2-6 d+9}=\frac{d-2}{3 d-9}} perfectly exemplifies this, with 'd' representing the variable we aim to solve for. What sets rational equations apart is the potential for extraneous solutions. These are solutions obtained through the algebraic process that, when plugged back into the original equation, render it undefined – typically due to division by zero. Therefore, the final step of verifying solutions is not just a formality, but a necessity in ensuring the correctness of our answer.

When tackling rational equations, a methodical approach is vital. Our initial step involves simplifying the equation. This usually means factoring polynomials in the denominators, identifying any common factors, and determining the least common denominator (LCD). This process not only makes the equation easier to manipulate but also highlights any values of the variable that would make the denominator zero, which are crucial to note as potential extraneous solutions. Next, we eliminate the fractions by multiplying both sides of the equation by the LCD. This transforms the rational equation into a polynomial equation, which we can then solve using standard algebraic techniques. These may include factoring, using the quadratic formula, or other methods appropriate to the degree of the polynomial. Finally, and most importantly, we must check each solution obtained against the original equation to confirm it does not result in division by zero. This verification step distinguishes rational equation solving from other algebraic problem-solving and is key to arriving at the correct solution set.

Step-by-Step Solution of the Equation

1. Factoring the Denominators

Our initial task is to factor the denominators of the given equation: d2d−6−3d2−6d+9=d−23d−9{\frac{d}{2 d-6}-\frac{3}{d^2-6 d+9}=\frac{d-2}{3 d-9}}. This allows us to identify common factors and simplify the equation. Let's break down each denominator:

  • 2d - 6: We can factor out a 2, resulting in 2(d - 3).
  • d² - 6d + 9: This is a quadratic expression. Recognizing it as a perfect square trinomial, we factor it into (d - 3)².
  • 3d - 9: Similarly, we factor out a 3, which gives us 3(d - 3).

Rewriting the equation with the factored denominators, we get:

d2(d−3)−3(d−3)2=d−23(d−3){\frac{d}{2(d - 3)}-\frac{3}{(d - 3)^2}=\frac{d-2}{3(d - 3)}}

This factorization step is crucial as it reveals the common factor (d - 3), which is essential for finding the least common denominator and identifying potential extraneous solutions. Factoring simplifies the equation and sets the stage for the next steps in solving it.

2. Finding the Least Common Denominator (LCD)

The least common denominator (LCD) is a cornerstone in solving rational equations. It is the smallest expression that is divisible by all the denominators in the equation. In our case, the denominators are 2(d - 3), (d - 3)², and 3(d - 3). To find the LCD, we consider each factor and its highest power present in any denominator:

  • The factor 2 appears in the first denominator.
  • The factor 3 appears in the third denominator.
  • The factor (d - 3) appears in all denominators, with the highest power being (d - 3)².

Therefore, the LCD is the product of these factors: 6(d - 3)². This LCD will be instrumental in eliminating the fractions from the equation, making it easier to solve. Identifying the correct LCD is vital for efficient and accurate solutions in rational equations.

3. Multiplying by the LCD

To eliminate the fractions in the equation d2(d−3)−3(d−3)2=d−23(d−3){\frac{d}{2(d - 3)}-\frac{3}{(d - 3)^2}=\frac{d-2}{3(d - 3)}}, we multiply both sides of the equation by the LCD, which we determined to be 6(d - 3)². This step transforms the rational equation into a more manageable polynomial equation.

Multiplying each term by the LCD, we get:

  • d2(d−3)×6(d−3)2=3d(d−3){\frac{d}{2(d - 3)} \times 6(d - 3)^2 = 3d(d - 3)}
  • 3(d−3)2×6(d−3)2=18{\frac{3}{(d - 3)^2} \times 6(d - 3)^2 = 18}
  • d−23(d−3)×6(d−3)2=2(d−2)(d−3){\frac{d-2}{3(d - 3)} \times 6(d - 3)^2 = 2(d - 2)(d - 3)}

Substituting these back into the equation, we have:

3d(d - 3) - 18 = 2(d - 2)(d - 3)

This equation is now free of fractions, making it easier to solve. The multiplication by the LCD has effectively cleared the denominators, setting us up for the next steps in finding the solution.

4. Simplifying and Solving the Equation

Now that we've cleared the fractions, we proceed to simplify and solve the resulting polynomial equation: 3d(d - 3) - 18 = 2(d - 2)(d - 3). This involves expanding the terms, combining like terms, and rearranging the equation into a standard form.

First, expand the terms:

  • 3d(d - 3) = 3d² - 9d
  • 2(d - 2)(d - 3) = 2(d² - 5d + 6) = 2d² - 10d + 12

Substituting these back into the equation, we get:

3d² - 9d - 18 = 2d² - 10d + 12

Next, we bring all terms to one side to set the equation to zero:

3d² - 9d - 18 - (2d² - 10d + 12) = 0

Simplify by combining like terms:

d² + d - 30 = 0

This is a quadratic equation, which we can solve by factoring. We look for two numbers that multiply to -30 and add to 1. These numbers are 6 and -5. Thus, we factor the quadratic as:

(d + 6)(d - 5) = 0

Setting each factor equal to zero gives us the potential solutions:

  • d + 6 = 0 => d = -6
  • d - 5 = 0 => d = 5

These are the candidate solutions to the equation. However, we must verify these solutions in the original equation to ensure they are not extraneous.

5. Checking for Extraneous Solutions

Checking for extraneous solutions is a critical step in solving rational equations. Extraneous solutions are values that satisfy the transformed equation but make the original equation undefined, typically due to division by zero. Our candidate solutions are d = -6 and d = 5. We must substitute each value back into the original equation d2d−6−3d2−6d+9=d−23d−9{\frac{d}{2 d-6}-\frac{3}{d^2-6 d+9}=\frac{d-2}{3 d-9}} to check for extraneous solutions.

Check d = -6:

Substitute d = -6 into the original equation:

−62(−6)−6−3(−6)2−6(−6)+9=−6−23(−6)−9{\frac{-6}{2(-6)-6}-\frac{3}{(-6)^2-6(-6)+9}=\frac{-6-2}{3(-6)-9}}

Simplify each term:

−6−18−336+36+9=−8−18−9{\frac{-6}{-18}-\frac{3}{36+36+9}=\frac{-8}{-18-9}}

13−381=−8−27{\frac{1}{3}-\frac{3}{81}=\frac{-8}{-27}}

13−127=827{\frac{1}{3}-\frac{1}{27}=\frac{8}{27}}

927−127=827{\frac{9}{27}-\frac{1}{27}=\frac{8}{27}}

827=827{\frac{8}{27}=\frac{8}{27}}

Since the equation holds true, d = -6 is a valid solution.

Check d = 5:

Substitute d = 5 into the original equation:

52(5)−6−3(5)2−6(5)+9=5−23(5)−9{\frac{5}{2(5)-6}-\frac{3}{(5)^2-6(5)+9}=\frac{5-2}{3(5)-9}}

Simplify each term:

510−6−325−30+9=315−9{\frac{5}{10-6}-\frac{3}{25-30+9}=\frac{3}{15-9}}

54−34=36{\frac{5}{4}-\frac{3}{4}=\frac{3}{6}}

24=12{\frac{2}{4}=\frac{1}{2}}

12=12{\frac{1}{2}=\frac{1}{2}}

Since the equation also holds true for d = 5, it is also a valid solution.

Therefore, both d = -6 and d = 5 are solutions to the original equation. This thorough checking process ensures that our solutions are valid and not extraneous.

Final Answer

After meticulously solving the rational equation and verifying the solutions, we arrive at the final answer. The solutions to the equation d2d−6−3d2−6d+9=d−23d−9{\frac{d}{2 d-6}-\frac{3}{d^2-6 d+9}=\frac{d-2}{3 d-9}} are d = -6 and d = 5. We obtained these solutions by factoring the denominators, finding the least common denominator, multiplying both sides of the equation by the LCD, simplifying the resulting polynomial equation, and then solving for d. Crucially, we checked both solutions in the original equation to confirm that they do not lead to division by zero, thus ensuring they are not extraneous.

Conclusion

Solving rational equations requires a blend of algebraic skill and careful attention to detail. The process involves factoring, finding the LCD, simplifying the equation, and solving for the variable. However, the most critical step is checking for extraneous solutions. This step distinguishes rational equation solving from other algebraic manipulations. By verifying each solution, we ensure the accuracy of our results. This comprehensive approach, demonstrated through the step-by-step solution of the given equation, equips you with the necessary tools to tackle similar problems with confidence.

Rational equations are a staple in algebra, and mastering their solution is essential for further studies in mathematics. The techniques discussed here – factoring, finding the LCD, solving the equation, and checking for extraneous solutions – are applicable to a wide range of rational equations. Remember, practice is key to proficiency. Work through various examples, focusing on the logical flow of each step and the importance of verification. This will solidify your understanding and enhance your problem-solving abilities in algebra and beyond.

A. The solution(s) is/are d = -6, 5.