Solving Sqrt(-6y+64) = -4 A Step-by-Step Guide

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Introduction

In this article, we will delve into the process of solving an algebraic equation for a specific variable. Specifically, we will tackle the equation βˆ’6y+64=βˆ’4\sqrt{-6y + 64} = -4, where vv is a real number. This problem falls under the category of mathematics, particularly algebra, and requires us to isolate the variable yy while adhering to the rules governing square roots and real numbers. The equation presents an interesting challenge due to the presence of a square root and a negative value on one side, which necessitates a careful and methodical approach to ensure we arrive at the correct solution, if one exists. This detailed exploration will enhance our understanding of algebraic manipulation and the properties of real numbers.

Understanding the Equation

To solve for vv, where vv is a real number, and we have the equation βˆ’6y+64=βˆ’4\sqrt{-6y + 64} = -4, our initial step involves understanding the equation's components and the mathematical principles that govern it. The equation contains a square root term, βˆ’6y+64\sqrt{-6y + 64}, which implies that the expression inside the square root, βˆ’6y+64-6y + 64, must be non-negative for the square root to yield a real number result. This is because the square root of a negative number is not a real number but an imaginary number. On the other side of the equation, we have βˆ’4-4, a negative number. This immediately raises a red flag because the square root of a real number is, by definition, non-negative. The principal square root, which is what we typically consider when solving equations, always returns the non-negative value. This means that the square root of any non-negative number will be either zero or a positive number. Therefore, it is crucial to recognize this inherent constraint of square roots before proceeding further.

The presence of the negative value on the right-hand side of the equation suggests that there might not be a real solution. However, to confirm this, we must rigorously follow the steps of algebraic manipulation and check for any inconsistencies. The process involves squaring both sides of the equation to eliminate the square root, solving for the variable yy, and then verifying whether the obtained solution(s) satisfy the original equation. This verification step is particularly important when dealing with square roots, as squaring both sides can sometimes introduce extraneous solutionsβ€”solutions that satisfy the transformed equation but not the original one. Understanding these nuances is essential for accurately solving equations involving radicals and avoiding common pitfalls.

Step-by-Step Solution

Now, let's proceed with the step-by-step solution to the equation βˆ’6y+64=βˆ’4\sqrt{-6y + 64} = -4. The first step in solving this equation is to eliminate the square root. To do this, we square both sides of the equation. Squaring both sides gives us (βˆ’6y+64)2=(βˆ’4)2(\sqrt{-6y + 64})^2 = (-4)^2. This simplifies to βˆ’6y+64=16-6y + 64 = 16. Now we have a linear equation in terms of yy, which is much easier to solve. The next step is to isolate the term containing yy. We can do this by subtracting 64 from both sides of the equation: βˆ’6y+64βˆ’64=16βˆ’64-6y + 64 - 64 = 16 - 64. This simplifies to βˆ’6y=βˆ’48-6y = -48. Finally, to solve for yy, we divide both sides of the equation by -6: βˆ’6yβˆ’6=βˆ’48βˆ’6\frac{-6y}{-6} = \frac{-48}{-6}. This gives us y=8y = 8.

However, finding a potential solution is only half the battle. With equations involving square roots, it is crucial to check whether the solution we obtained is a valid solution or an extraneous one. An extraneous solution is a value that satisfies the transformed equation (in this case, after squaring both sides) but does not satisfy the original equation. To check our solution, we substitute y=8y = 8 back into the original equation: βˆ’6y+64=βˆ’4\sqrt{-6y + 64} = -4. Substituting y=8y = 8, we get βˆ’6(8)+64=βˆ’4\sqrt{-6(8) + 64} = -4, which simplifies to βˆ’48+64=βˆ’4\sqrt{-48 + 64} = -4. Further simplification gives us 16=βˆ’4\sqrt{16} = -4. The square root of 16 is 4, so we have 4=βˆ’44 = -4. This statement is clearly false. Therefore, y=8y = 8 is not a solution to the original equation. This result underscores the importance of the verification step when solving equations involving radicals.

Verification and Extraneous Solutions

The verification step is critical when solving equations, especially those involving radicals like square roots. As we saw in the previous section, squaring both sides of an equation can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. To illustrate this further, let's revisit the equation βˆ’6y+64=βˆ’4\sqrt{-6y + 64} = -4 and the solution we obtained, y=8y = 8. When we substituted y=8y = 8 back into the original equation, we found that βˆ’6(8)+64=16=4\sqrt{-6(8) + 64} = \sqrt{16} = 4, which is not equal to -4. This discrepancy confirms that y=8y = 8 is an extraneous solution. The reason for this lies in the fundamental property of square roots: the principal square root of a number is always non-negative. In other words, x\sqrt{x} is defined to be the non-negative value that, when squared, gives xx. Therefore, 16\sqrt{16} is 4, not -4.

The extraneous solution arises because squaring both sides of the equation effectively eliminates the distinction between positive and negative roots. The equation (βˆ’6y+64)2=(βˆ’4)2(\sqrt{-6y + 64})^2 = (-4)^2 is equivalent to βˆ’6y+64=16-6y + 64 = 16, which would also be obtained if we had the equation βˆ’6y+64=4\sqrt{-6y + 64} = 4. This means that the transformed equation encompasses solutions for both the original equation and its counterpart with the positive square root. To avoid mistakenly accepting extraneous solutions, it is always necessary to substitute the potential solutions back into the original equation and check for validity. This step ensures that the solutions we obtain are genuine and not artifacts of the algebraic manipulations we performed.

Conclusion and Final Answer

In conclusion, after solving the equation βˆ’6y+64=βˆ’4\sqrt{-6y + 64} = -4 and rigorously verifying our solution, we have determined that there is no real number solution for yy. We initially found y=8y = 8 as a potential solution by squaring both sides, isolating yy, and performing the necessary algebraic manipulations. However, upon substituting y=8y = 8 back into the original equation, we discovered that it led to the contradiction 4=βˆ’44 = -4. This contradiction indicates that y=8y = 8 is an extraneous solution, meaning it satisfies the transformed equation but not the original one. The fundamental reason for the absence of a real solution lies in the nature of square roots. The principal square root of a real number is always non-negative, so βˆ’6y+64\sqrt{-6y + 64} cannot be equal to -4, a negative number.

This problem highlights the importance of not only performing algebraic manipulations correctly but also understanding the underlying mathematical principles and constraints. The verification step is crucial in solving equations involving radicals, as it helps us identify and eliminate extraneous solutions. By carefully checking our solutions, we can ensure that we arrive at the correct answer. In this case, the correct answer is that there is no real solution for the equation βˆ’6y+64=βˆ’4\sqrt{-6y + 64} = -4. Therefore, the final answer is no solution.

Final Answer: No solution