Solving Y + 6 = √(2y² + 72) A Step-by-Step Guide
This article delves into the step-by-step solution of the equation y + 6 = √(2y² + 72). We will explore the algebraic manipulations required to isolate y and arrive at the correct solution(s). This type of equation, involving a square root, often necessitates careful consideration of potential extraneous solutions, which we will address thoroughly. Understanding how to solve such equations is fundamental in algebra and crucial for various applications in mathematics and other scientific disciplines.
Step-by-Step Solution
Let's break down the solution process into manageable steps:
1. Isolate the Square Root
The initial equation is: y + 6 = √(2y² + 72).
In this case, the square root term is already isolated on one side of the equation, which simplifies our first step. This is a crucial starting point because we need to eliminate the square root to work with the equation more easily. When encountering equations with radicals, the first goal should always be to isolate the radical term. This sets the stage for the next step, which involves squaring both sides of the equation. By isolating the square root, we ensure that the squaring operation will effectively remove the radical, allowing us to proceed with solving for the variable.
2. Square Both Sides
To eliminate the square root, we square both sides of the equation:
(y + 6)² = (√(2y² + 72))²
Squaring both sides is a key algebraic technique used to eliminate square roots. However, it's important to remember that this operation can sometimes introduce extraneous solutions. These are solutions that satisfy the transformed equation but not the original equation. Therefore, it's crucial to check all potential solutions in the original equation to ensure they are valid. Squaring both sides allows us to transform the equation into a more manageable form, typically a polynomial equation, which we can then solve using standard algebraic methods. In this specific case, squaring both sides will result in a quadratic equation, which we will then solve for y. This step is fundamental in our quest to find the values of y that satisfy the original equation.
3. Expand and Simplify
Expanding the left side gives us:
y² + 12y + 36 = 2y² + 72
Now, let's simplify the equation by moving all terms to one side to set the equation to zero. Subtract y², 12y, and 36 from both sides:
0 = 2y² - y² - 12y + 72 - 36
This simplifies to:
0 = y² - 12y + 36
Expanding and simplifying are crucial steps in solving equations. Expanding, in this context, involves removing parentheses by applying the distributive property or by using identities such as (a + b)² = a² + 2ab + b². Simplification involves combining like terms and rearranging the equation into a standard form, such as a polynomial equation. The goal of these steps is to transform the equation into a more manageable form that is easier to solve. In our case, expanding the squared term and then simplifying by collecting like terms leads us to a quadratic equation. This quadratic equation is now in a form that we can readily solve using methods such as factoring, completing the square, or the quadratic formula. The accuracy of these steps is paramount, as any error in expansion or simplification will propagate through the rest of the solution process.
4. Solve the Quadratic Equation
The equation y² - 12y + 36 = 0 is a quadratic equation. We can solve this by factoring:
(y - 6)(y - 6) = 0
This gives us:
(y - 6)² = 0
Therefore, y = 6.
Solving the quadratic equation is a pivotal step in finding the solution to our original problem. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. The choice of method often depends on the specific form of the equation. In this case, the quadratic equation y² - 12y + 36 = 0 is easily factorable, which makes factoring the most efficient method. Factoring involves expressing the quadratic expression as a product of two binomials. By setting each factor equal to zero, we can find the roots, or solutions, of the equation. In this specific instance, the quadratic equation factors to (y - 6)(y - 6) = 0, which means that the equation has a repeated root at y = 6. Recognizing and applying the appropriate method for solving quadratic equations is a fundamental skill in algebra and is crucial for solving a wide range of mathematical problems.
5. Check for Extraneous Solutions
Since we squared both sides of the equation, we need to check if our solution is extraneous. Substitute y = 6 back into the original equation:
6 + 6 = √(2(6)² + 72)
12 = √(2(36) + 72)
12 = √(72 + 72)
12 = √144
12 = 12
The solution y = 6 satisfies the original equation.
Checking for extraneous solutions is a critical step when solving equations that involve squaring both sides or other operations that can introduce solutions that do not satisfy the original equation. Extraneous solutions arise because the operation of squaring both sides can sometimes create a new equation with a broader set of solutions than the original equation. To check for extraneous solutions, each potential solution must be substituted back into the original equation to verify that it makes the equation true. If a potential solution does not satisfy the original equation, it is deemed extraneous and must be discarded. In our case, we found a potential solution of y = 6. By substituting this value back into the original equation y + 6 = √(2y² + 72), we verified that it indeed satisfies the equation, confirming that y = 6 is a valid solution. This step is essential to ensure the accuracy of the final solution set.
Final Answer
Therefore, the solution to the equation y + 6 = √(2y² + 72) is:
y = 6
This comprehensive guide has walked you through the process of solving the equation y + 6 = √(2y² + 72). We have emphasized the importance of each step, from isolating the square root to checking for extraneous solutions. By understanding these principles, you can confidently tackle similar algebraic problems.
