Solving 5x = √(30 - 5x) A Step-by-Step Guide
In this article, we will delve into the process of solving the equation 5x = √(30 - 5x) for x. This equation is a blend of algebraic expressions and a square root, requiring a methodical approach to isolate the variable and determine its value. We will explore the steps involved, the potential pitfalls of extraneous solutions, and how to verify the solutions we obtain. Understanding how to solve equations of this nature is fundamental in algebra and provides a strong foundation for more advanced mathematical concepts. As we proceed, we'll emphasize clarity and precision, ensuring that each step is logically sound and contributes to the final solution. Whether you're a student grappling with algebra or simply someone with an interest in mathematical problem-solving, this guide will offer valuable insights and practical techniques.
Step-by-Step Solution
To effectively solve the equation 5x = √(30 - 5x), we will follow a systematic approach, breaking down the problem into manageable steps. This will not only help us arrive at the correct solution but also provide a clear understanding of the underlying mathematical principles. Each step is crucial, and we will explain the rationale behind it, ensuring that the process is transparent and easy to follow. Our goal is to not just find the solution but to also equip you with the skills to tackle similar problems in the future.
1. Squaring Both Sides
The initial step in solving the equation 5x = √(30 - 5x) involves eliminating the square root. To achieve this, we square both sides of the equation. This operation is based on the principle that if two quantities are equal, then their squares are also equal. Squaring both sides allows us to transform the equation into a more manageable form, specifically a quadratic equation. However, it's crucial to recognize that squaring both sides can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. Therefore, we must be vigilant about verifying our solutions later in the process.
Squaring the left side, (5x), yields 25x². Squaring the right side, √(30 - 5x), eliminates the square root, resulting in 30 - 5x. Thus, the equation becomes:
25x² = 30 - 5x
This transformation is a significant step forward, as it converts the original equation into a quadratic equation, which we can then solve using standard techniques.
2. Rearranging into a Quadratic Equation
Now that we have the equation 25x² = 30 - 5x, the next step is to rearrange it into the standard form of a quadratic equation, which is ax² + bx + c = 0. This form is essential because it allows us to apply various methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. Rearranging the terms ensures that all terms are on one side of the equation, with the other side being zero. This standardization is a critical step in solving quadratic equations efficiently and accurately.
To rearrange the equation 25x² = 30 - 5x, we need to move all terms to one side. We can do this by adding 5x to both sides and subtracting 30 from both sides. This gives us:
25x² + 5x - 30 = 0
This equation is now in the standard quadratic form, where a = 25, b = 5, and c = -30. Having the equation in this form sets the stage for the next step, which involves simplifying the equation to make it easier to solve.
3. Simplifying the Quadratic Equation
The quadratic equation we obtained, 25x² + 5x - 30 = 0, can be simplified to make it easier to solve. Simplification involves dividing all terms of the equation by a common factor, which reduces the coefficients and makes the equation more manageable. This step is crucial because it can significantly reduce the complexity of the equation, making it easier to factor or apply the quadratic formula. In this case, we observe that all coefficients (25, 5, and -30) are divisible by 5. Dividing by the greatest common factor simplifies the equation without changing its solutions.
Dividing each term of the equation 25x² + 5x - 30 = 0 by 5, we get:
(25x² / 5) + (5x / 5) - (30 / 5) = 0
This simplifies to:
5x² + x - 6 = 0
This simplified quadratic equation is much easier to work with than the original. The next step will involve solving this simplified equation, which we can do through factoring or by using the quadratic formula.
4. Solving the Simplified Quadratic Equation
With the simplified quadratic equation 5x² + x - 6 = 0 in hand, we now proceed to solve for x. There are several methods to solve a quadratic equation, including factoring, completing the square, and using the quadratic formula. In this case, we will attempt to solve by factoring, as it is often the quickest method when applicable. Factoring involves expressing the quadratic expression as a product of two binomials. If factoring proves difficult or impossible, we can always resort to the quadratic formula, which provides a general solution for any quadratic equation.
To factor 5x² + x - 6 = 0, we look for two numbers that multiply to (5 * -6) = -30 and add up to 1 (the coefficient of x). These numbers are 6 and -5. We can rewrite the middle term using these numbers:
5x² - 5x + 6x - 6 = 0
Now we factor by grouping:
5x(x - 1) + 6(x - 1) = 0
We can factor out the common binomial (x - 1):
(5x + 6)(x - 1) = 0
This gives us two possible solutions for x:
5x + 6 = 0 or x - 1 = 0
Solving these equations, we get:
x = -6/5 or x = 1
These are the potential solutions to our quadratic equation. However, it's essential to remember that we squared both sides of the original equation, which may have introduced extraneous solutions. Therefore, we must verify these solutions in the original equation.
5. Verifying the Solutions
After obtaining potential solutions, it is crucial to verify them in the original equation. This step is essential because squaring both sides of an equation, as we did in the beginning, can introduce extraneous solutions. Extraneous solutions are values that satisfy the transformed equation (in our case, the quadratic equation) but do not satisfy the original equation. They arise because the squaring operation can make a negative quantity appear positive, thus altering the solution set. Verifying the solutions ensures that we only accept those values that truly make the original equation true.
Our original equation is 5x = √(30 - 5x), and our potential solutions are x = -6/5 and x = 1. We will substitute each value back into the original equation to check if it holds true.
Verifying x = -6/5
Substituting x = -6/5 into the original equation:
5(-6/5) = √(30 - 5(-6/5))
Simplifying the left side:
-6 = √(30 + 6)
-6 = √36
-6 = 6
This statement is false. Therefore, x = -6/5 is an extraneous solution and is not a valid solution to the original equation.
Verifying x = 1
Substituting x = 1 into the original equation:
5(1) = √(30 - 5(1))
Simplifying the left side:
5 = √(30 - 5)
5 = √25
5 = 5
This statement is true. Therefore, x = 1 is a valid solution to the original equation.
Final Solution
After careful verification, we have determined that the only valid solution to the equation 5x = √(30 - 5x) is x = 1. The solution x = -6/5 was found to be an extraneous solution, which does not satisfy the original equation. This highlights the importance of the verification step when solving equations involving radicals or rational exponents. The extraneous solution arose due to the squaring of both sides of the equation, which introduced a potential solution that did not exist in the original equation's domain. Our step-by-step approach, from isolating the radical to verifying the solutions, has ensured that we arrive at the correct answer with confidence.
Therefore, the solution to the equation 5x = √(30 - 5x) is:
x = 1
This concludes our comprehensive exploration of solving the equation. We have covered each step in detail, explaining the reasoning and potential pitfalls along the way. This method not only provides the solution but also enhances understanding of the underlying mathematical principles.