Solving The Equation T-√(3t+6)=-2 A Step-by-Step Guide

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Introduction: Tackling the Square Root Equation

In the realm of mathematics, solving equations is a fundamental skill, and equations involving radicals, particularly square roots, often present a unique challenge. This article delves into the intricacies of solving the equation $t-\sqrt{3t+6}=-2$, providing a step-by-step guide and exploring the underlying concepts. We will dissect the equation, isolate the radical term, eliminate the square root, and ultimately arrive at the solution(s). Understanding the process of solving such equations is crucial for students, educators, and anyone with a passion for mathematics. This exploration will not only provide the solution but also enhance your problem-solving skills and deepen your appreciation for the elegance of algebraic manipulation. To successfully navigate this equation, we'll employ a combination of algebraic techniques, including isolating the radical, squaring both sides, and solving the resulting quadratic equation. Along the way, we'll emphasize the importance of checking for extraneous solutions, which can arise when squaring both sides of an equation. By the end of this discussion, you'll have a comprehensive understanding of how to solve this specific equation and be better equipped to tackle similar problems in the future. This methodical approach will not only lead us to the correct answers but also illuminate the mathematical principles at play, making the process both educational and engaging.

Step 1: Isolating the Radical Term

The first crucial step in solving this equation is to isolate the radical term. This means rearranging the equation so that the square root term, $\sqrt{3t+6}$, is by itself on one side of the equation. To achieve this, we add $\sqrt{3t+6}$ to both sides and add 2 to both sides of the equation:

t−3t+6=−2t-\sqrt{3t+6} = -2

t+2=3t+6t + 2 = \sqrt{3t+6}

By isolating the radical, we set the stage for the next step, which involves eliminating the square root. This rearrangement is a critical step in simplifying the equation and making it more amenable to algebraic manipulation. This step is not just about moving terms around; it's about strategically positioning the radical term so that it can be effectively dealt with. The isolation of the radical is a common technique in solving equations involving radicals, and it's a skill that's applicable to a wide range of problems. It allows us to transform a complex equation into a more manageable form, paving the way for further simplification and ultimately, the solution. The beauty of this step lies in its ability to declutter the equation, bringing the essential element – the radical – into focus, ready for the next phase of the solution process. With the radical now isolated, we can proceed with confidence, knowing that we've laid a solid foundation for the subsequent steps.

Step 2: Eliminating the Square Root

With the radical term isolated, the next logical step is to eliminate the square root. To accomplish this, we square both sides of the equation. This operation is based on the principle that if two quantities are equal, their squares are also equal. Squaring both sides of the equation $t + 2 = \sqrt{3t+6}$ gives us:

(t+2)2=(3t+6)2(t + 2)^2 = (\sqrt{3t+6})^2

Expanding the left side and simplifying the right side, we get:

t2+4t+4=3t+6t^2 + 4t + 4 = 3t + 6

This step is crucial because it transforms the equation from one involving a square root into a more familiar quadratic equation. Squaring both sides is a powerful technique for eliminating radicals, but it's also important to recognize that this operation can sometimes introduce extraneous solutions. Therefore, it's essential to check our solutions at the end of the process to ensure they are valid. The act of squaring both sides effectively undoes the square root, allowing us to work with a polynomial equation that is generally easier to solve. This transformation is a cornerstone of solving radical equations, and it highlights the interconnectedness of different algebraic concepts. By understanding how to manipulate equations in this way, we can unlock solutions that might otherwise remain hidden. The resulting quadratic equation now presents a clear path forward, allowing us to apply standard techniques for solving polynomial equations.

Step 3: Solving the Quadratic Equation

Now that we have a quadratic equation, $t^2 + 4t + 4 = 3t + 6$, we need to solve it for t. To do this, we first rearrange the equation to the standard quadratic form, which is $ax^2 + bx + c = 0$. Subtracting 3t and 6 from both sides, we get:

t2+4t+4−3t−6=0t^2 + 4t + 4 - 3t - 6 = 0

t2+t−2=0t^2 + t - 2 = 0

Next, we can factor the quadratic expression:

(t+2)(t−1)=0(t + 2)(t - 1) = 0

This gives us two potential solutions for t:

t+2=0⇒t=−2t + 2 = 0 \Rightarrow t = -2

t−1=0⇒t=1t - 1 = 0 \Rightarrow t = 1

The process of solving a quadratic equation is a fundamental skill in algebra, and it's a crucial step in finding the solutions to our original radical equation. By rearranging the equation into standard form and then factoring, we were able to identify the values of t that satisfy the equation. However, it's important to remember that squaring both sides of an equation can sometimes introduce extraneous solutions, so we must proceed to the next step to verify our solutions. The ability to factor quadratic expressions is a powerful tool in solving equations, and it's a technique that's widely applicable in mathematics and other fields. The solutions we've found are potential answers to our original problem, but we must now subject them to a rigorous check to ensure their validity. This step highlights the importance of carefulness and attention to detail in mathematical problem-solving, ensuring that we arrive at the correct and meaningful answers.

Step 4: Checking for Extraneous Solutions

An essential step in solving equations involving radicals is to check for extraneous solutions. These are solutions that satisfy the transformed equation (after squaring) but do not satisfy the original equation. We must substitute each potential solution back into the original equation, $t-\sqrt{3t+6}=-2$, to verify its validity.

Checking t = -2:

−2−3(−2)+6=−2−0=−2-2 - \sqrt{3(-2) + 6} = -2 - \sqrt{0} = -2

Since -2 = -2, t = -2 is a valid solution.

Checking t = 1:

1−3(1)+6=1−9=1−3=−21 - \sqrt{3(1) + 6} = 1 - \sqrt{9} = 1 - 3 = -2

Since -2 = -2, t = 1 is also a valid solution.

This step is not merely a formality; it's a critical safeguard against incorrect answers. Extraneous solutions can arise because the operation of squaring both sides of an equation can introduce solutions that don't actually satisfy the original equation. The act of substituting our potential solutions back into the original equation is a process of verification, ensuring that our solutions are not just mathematical artifacts but are genuine solutions to the problem at hand. The fact that both of our potential solutions checked out in this case is a testament to the carefulness of our work, but it's a reminder that this step should never be skipped. The process of checking for extraneous solutions reinforces the importance of mathematical rigor and the need to validate our findings. By performing this check, we can confidently assert that we have found the true solutions to the equation.

Conclusion: The Solutions and the Process

In conclusion, by meticulously following the steps of isolating the radical, eliminating the square root, solving the quadratic equation, and checking for extraneous solutions, we have successfully determined that the solutions to the equation $t-\sqrt{3t+6}=-2$ are t = -2 and t = 1. This journey through the equation highlights the importance of a systematic approach to problem-solving in mathematics. Each step, from isolating the radical to checking for extraneous solutions, plays a crucial role in arriving at the correct answer. The process of solving radical equations is not just about finding the numbers that make the equation true; it's about understanding the underlying principles and applying them with precision. The skills we've honed in solving this equation are transferable to a wide range of mathematical problems, and the understanding we've gained will serve us well in future endeavors. The solutions we've found are not just numbers; they are the culmination of a careful and thoughtful process, a testament to the power of mathematical reasoning. This exploration has not only provided us with the answers but has also enriched our understanding of the mathematical landscape, preparing us for further challenges and discoveries.