Solving (y+6)^2=4 A Step-by-Step Guide
In this article, we will delve into the step-by-step process of solving the equation (y+6)^2=4, where y represents a real number. Our goal is to simplify the answer as much as possible and present the solution(s) in a clear and understandable manner. Whether you're a student grappling with algebraic equations or simply seeking to enhance your mathematical prowess, this guide will provide you with the necessary tools and insights to tackle similar problems with confidence.
Understanding the Equation
Before we embark on the solution, let's take a moment to dissect the equation (y+6)^2=4. At its core, this is a quadratic equation disguised in a slightly different form. The presence of the squared term, (y+6)^2, indicates that we're dealing with a quadratic relationship. However, instead of the standard form of a quadratic equation (ax^2 + bx + c = 0), we have a squared binomial on one side and a constant on the other. This particular structure lends itself to a straightforward solution using the square root property.
The square root property states that if x^2 = a, then x = ±√a. In essence, this property allows us to undo the squaring operation by taking the square root of both sides of the equation. However, it's crucial to remember that when taking the square root, we must consider both the positive and negative roots, as both will satisfy the original equation. This is because squaring either a positive or a negative number yields a positive result. For instance, both 2 and -2, when squared, result in 4.
In our equation, (y+6)^2=4, the term (y+6) plays the role of x in the square root property, and 4 plays the role of a. Therefore, applying the square root property will be our first step in unraveling the solution. Let's move on to the step-by-step solution process.
Step-by-Step Solution
Now, let's embark on the journey of solving the equation (y+6)^2=4. We will meticulously walk through each step, ensuring clarity and understanding along the way. Our approach will be guided by the square root property, which, as we discussed earlier, provides a direct path to isolating the variable y.
Step 1: Applying the Square Root Property
The first and most crucial step is to apply the square root property to both sides of the equation. As mentioned before, this property states that if x^2 = a, then x = ±√a. In our case, x corresponds to (y+6), and a corresponds to 4. Therefore, we can rewrite the equation as:
y + 6 = ±√4
This step is paramount because it effectively undoes the squaring operation, bringing us closer to isolating y. The ± symbol signifies that we need to consider both the positive and negative square roots of 4, which is a critical aspect of solving quadratic equations.
Step 2: Evaluating the Square Root
The next step involves evaluating the square root of 4. We know that the square root of 4 is 2, since 2 multiplied by itself equals 4. However, as we discussed earlier, we must consider both the positive and negative roots. Therefore, we have:
y + 6 = ±2
This step simplifies the equation further, replacing the square root symbol with its numerical value. Now, we have two separate equations to solve, one for the positive root and one for the negative root.
Step 3: Isolating y
To isolate y, we need to get it by itself on one side of the equation. In both equations, y is being added to 6. To undo this addition, we will subtract 6 from both sides of each equation. This will give us two separate solutions for y:
Case 1: Positive Root
y + 6 = +2 y = 2 - 6 y = -4
Case 2: Negative Root
y + 6 = -2 y = -2 - 6 y = -8
This step is the culmination of our algebraic manipulation, leading us to the individual solutions for y. We have successfully isolated y in both cases, revealing the two values that satisfy the original equation.
Step 4: Presenting the Solution
Finally, we present our solutions for y. We have found two values that satisfy the equation (y+6)^2=4: -4 and -8. Therefore, the solution can be expressed as:
y = -4, -8
This is the simplified solution to the equation. We have successfully navigated the steps, from understanding the equation to applying the square root property and isolating the variable. This systematic approach can be applied to solving a variety of similar equations.
Verification of the Solutions
To ensure the accuracy of our solutions, it's always a good practice to verify them. We can do this by substituting each solution back into the original equation and checking if the equation holds true. Let's verify our solutions, y = -4 and y = -8.
Verification for y = -4
Substituting y = -4 into the original equation (y+6)^2=4, we get:
(-4 + 6)^2 = 4 (2)^2 = 4 4 = 4
The equation holds true, confirming that y = -4 is indeed a valid solution.
Verification for y = -8
Now, let's substitute y = -8 into the original equation:
(-8 + 6)^2 = 4 (-2)^2 = 4 4 = 4
Again, the equation holds true, validating y = -8 as a solution. The verification process provides a safety net, ensuring that our algebraic manipulations have led us to the correct answers. It reinforces our confidence in the solutions we've obtained.
Alternative Methods for Solving Quadratic Equations
While the square root property provided a direct and efficient method for solving (y+6)^2=4, it's worth noting that other methods exist for solving quadratic equations in general. Understanding these alternative approaches can broaden your problem-solving toolkit and provide valuable insights into the nature of quadratic equations.
1. Expanding and Factoring
One approach is to expand the squared binomial, rewrite the equation in standard quadratic form (ay^2 + by + c = 0), and then attempt to factor the quadratic expression. This method relies on the ability to recognize and apply factoring techniques. Let's illustrate this with our equation:
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Expand the squared binomial:
(y+6)^2 = y^2 + 12y + 36
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Rewrite the equation:
y^2 + 12y + 36 = 4
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Move the constant term to the left side:
y^2 + 12y + 32 = 0
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Factor the quadratic expression:
(y + 4)(y + 8) = 0
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Set each factor to zero and solve:
y + 4 = 0 => y = -4 y + 8 = 0 => y = -8
As you can see, this method yields the same solutions as the square root property. However, factoring can be more challenging when the quadratic expression is not easily factorable.
2. Quadratic Formula
The quadratic formula is a universal tool for solving quadratic equations. It can be applied to any quadratic equation in the form ay^2 + by + c = 0, regardless of whether it's factorable or not. The formula is given by:
y = (-b ± √(b^2 - 4ac)) / 2a
To use the quadratic formula, we first need to rewrite our equation in standard form, as we did in the factoring method:
y^2 + 12y + 32 = 0
Now, we can identify the coefficients: a = 1, b = 12, and c = 32. Plugging these values into the quadratic formula, we get:
y = (-12 ± √(12^2 - 4 * 1 * 32)) / (2 * 1) y = (-12 ± √(144 - 128)) / 2 y = (-12 ± √16) / 2 y = (-12 ± 4) / 2
This gives us two solutions:
y = (-12 + 4) / 2 = -4 y = (-12 - 4) / 2 = -8
Again, the quadratic formula confirms our solutions. While it might seem more complex than the square root property in this specific case, the quadratic formula is a powerful tool for solving any quadratic equation.
Conclusion
In this comprehensive guide, we have successfully solved the equation (y+6)^2=4, demonstrating a step-by-step approach using the square root property. We also verified our solutions to ensure their accuracy. Furthermore, we explored alternative methods for solving quadratic equations, including factoring and the quadratic formula, highlighting the versatility of mathematical tools.
By understanding the underlying principles and mastering these techniques, you can confidently tackle a wide range of quadratic equations and enhance your problem-solving skills in mathematics. Remember, practice is key to solidifying your understanding and building your mathematical intuition.
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