Solving For B Values In The Equation 3(2b+3)²=36

This article delves into the realm of algebraic equations, specifically focusing on solving for the unknown variable b in the equation 3(2b+3)²=36. We will systematically break down the equation, step by step, to isolate b and determine the values that satisfy the given condition. Understanding the techniques used to solve such equations is crucial for mastering algebraic manipulation and problem-solving in mathematics. This exploration will not only provide the solutions for b but also illuminate the underlying principles of equation solving. We will discuss the importance of order of operations, the concept of inverse operations, and the application of the square root property. Furthermore, we will examine how the structure of the equation influences the solution process, highlighting the role of the squared term and the constant coefficients. By the end of this article, you will have a comprehensive understanding of how to tackle similar algebraic problems with confidence and accuracy. The core concept we will be using is to simplify the equation by performing inverse operations on both sides, ultimately isolating the variable b. This involves dividing by constants, taking square roots, and applying the distributive property in reverse. This systematic approach will ensure that we arrive at the correct solutions while maintaining the equality of the equation throughout the process. Mastering these techniques is essential for success in higher-level mathematics and related fields.

H2: A Step-by-Step Solution

Let's embark on the journey of solving the equation 3(2b+3)²=36. Our primary objective is to isolate b, and we'll achieve this by meticulously applying algebraic operations to both sides of the equation. To begin, we notice that the entire squared term (2b+3)² is being multiplied by 3. To undo this multiplication, we perform the inverse operation, which is division. Dividing both sides of the equation by 3, we get (2b+3)²=12. This simplification has brought us closer to isolating b. The next hurdle is the squared term. To eliminate the square, we employ the square root property. Taking the square root of both sides, we must remember to consider both the positive and negative roots, as squaring either a positive or negative number results in a positive value. Thus, we obtain 2b+3=±√12. Now, we can simplify √12 as √(43), which equals 2√3. So, our equation becomes 2b*+3=±2√3. The equation now presents us with two separate cases to consider: 2b+3=2√3 and 2b+3=-2√3. Let's tackle the first case. To isolate b, we subtract 3 from both sides, resulting in 2b=-3+2√3. Then, we divide both sides by 2 to get b=(-3+2√3)/2. This is one solution for b. Now, let's address the second case: 2b+3=-2√3. Again, we subtract 3 from both sides, yielding 2b=-3-2√3. Finally, we divide both sides by 2 to get b=(-3-2√3)/2. This is our second solution for b. Therefore, the two values of b that satisfy the equation are b=(-3+2√3)/2 and b=(-3-2√3)/2. This step-by-step approach demonstrates the importance of careful algebraic manipulation and the consideration of all possible solutions when dealing with square roots.

H2: Detailed Breakdown of the Solution Process

The solution process for the equation 3(2b+3)²=36 involves a series of algebraic manipulations designed to isolate the variable b. Each step is crucial in ensuring we arrive at the correct solutions. Initially, we encounter the equation 3(2b+3)²=36. The presence of the constant 3 multiplying the squared term suggests our first move: division. Dividing both sides by 3 simplifies the equation to (2b+3)²=12. This step effectively removes the coefficient outside the parentheses, bringing us closer to the core of the expression containing b. Next, we address the squared term. To undo the squaring operation, we take the square root of both sides. A critical point here is to remember that the square root operation yields both positive and negative results. This is because both a positive and a negative number, when squared, will produce a positive result. Thus, taking the square root of (2b+3)² gives us 2b+3=±√12. Simplifying the square root of 12, we recognize that 12 can be factored as 43, and the square root of 4 is 2. This allows us to rewrite √12 as 2√3, leading to the equation 2b*+3=±2√3. At this juncture, we have two separate linear equations to solve: 2b+3=2√3 and 2b+3=-2√3. We solve each equation independently. For the first equation, 2b+3=2√3, we subtract 3 from both sides to get 2b=-3+2√3. Then, we divide by 2 to isolate b, resulting in b=(-3+2√3)/2. For the second equation, 2b+3=-2√3, we similarly subtract 3 from both sides to get 2b=-3-2√3. Dividing by 2 gives us b=(-3-2√3)/2. Therefore, the two solutions for b are b=(-3+2√3)/2 and b=(-3-2√3)/2. This detailed breakdown highlights the importance of considering both positive and negative roots when dealing with squared terms and the methodical application of inverse operations to isolate the variable. The algebraic techniques used here are fundamental to solving a wide range of equations in mathematics.

H2: Verifying the Solutions

To ensure the accuracy of our solutions, it's crucial to verify them by substituting them back into the original equation, 3(2b+3)²=36. This process confirms that the values we've found for b indeed satisfy the equation. Let's first consider the solution b=(-3+2√3)/2. Substituting this value into the equation, we get 3[2((-3+2√3)/2)+3]²=36. Simplifying the expression inside the brackets, the 2 in the numerator and denominator cancel out, leaving us with 3[-3+2√3+3]²=36. The -3 and +3 cancel each other, resulting in 3(2√3)²=36. Squaring 2√3 gives us (2√3)²=43=12. Thus, the equation becomes 3(12)=36, which simplifies to 36=36. This confirms that b=(-3+2√3)/2 is indeed a solution. Now, let's verify the second solution, b=(-3-2√3)/2. Substituting this value into the equation, we get 3[2((-3-2√3)/2)+3]²=36. Again, the 2 in the numerator and denominator cancel out, giving us 3[-3-2√3+3]²=36. The -3 and +3 cancel each other, leaving us with 3(-2√3)²=36. Squaring -2√3 gives us (-2√3)²=43=12. Thus, the equation becomes 3(12)=36, which simplifies to 36=36. This confirms that b=(-3-2√3)/2 is also a solution. By verifying both solutions, we gain confidence in our solution process and ensure that we have accurately determined the values of b that satisfy the equation. This verification step is a vital part of the problem-solving process in mathematics and should be employed whenever possible.

H2: Connecting to Broader Mathematical Concepts

Solving the equation 3(2b+3)²=36 not only provides us with specific solutions for b but also connects to broader mathematical concepts and principles. The equation itself is a quadratic equation in disguise. While it's not immediately apparent in the standard form ax²+bx+c=0, the presence of the squared term (2b+3)² indicates its quadratic nature. Solving this equation involves techniques that are applicable to solving a wide range of quadratic equations. One key concept is the square root property, which we utilized to eliminate the squared term. This property states that if x²=a, then x=±√a. This is a fundamental tool in solving equations where a variable or an expression is squared. Another important concept is the order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). We followed this order in reverse when isolating b, first addressing multiplication and division before dealing with addition and subtraction. Furthermore, the equation highlights the importance of considering both positive and negative solutions when taking square roots. This is because both the positive and negative square roots, when squared, yield the same positive result. The solutions we obtained, b=(-3+2√3)/2 and b=(-3-2√3)/2, are irrational numbers due to the presence of the √3 term. This connects to the broader concept of number systems, which includes rational, irrational, real, and complex numbers. In summary, solving the equation 3(2b+3)²=36 provides a practical application of several fundamental mathematical concepts, including quadratic equations, the square root property, order of operations, and number systems. Understanding these connections enhances our mathematical proficiency and problem-solving abilities.

H2: Conclusion: The Values of b That Satisfy the Equation

In conclusion, after a detailed and systematic solution process, we have determined the values of b that satisfy the equation 3(2b+3)²=36. The solutions are b=(-3+2√3)/2 and b=(-3-2√3)/2. These values were obtained by carefully applying algebraic manipulations, including dividing both sides of the equation by 3, taking the square root of both sides (remembering to consider both positive and negative roots), and isolating b. We also verified these solutions by substituting them back into the original equation, confirming their validity. The solution process not only provides the answers but also illustrates the application of fundamental algebraic principles, such as the square root property and the order of operations. Furthermore, it connects to broader mathematical concepts like quadratic equations and number systems. Mastering the techniques used to solve this type of equation is essential for success in algebra and beyond. The ability to systematically break down an equation, apply appropriate operations, and verify the results is a crucial skill in mathematical problem-solving. This exploration has highlighted the importance of precision, attention to detail, and a thorough understanding of algebraic principles in arriving at accurate solutions. The values of b we found represent the points where the expression 3(2b+3)² equals 36. This understanding can be further extended to graphical representations of equations and their solutions. By understanding the solution process and the underlying mathematical concepts, we can confidently tackle similar problems and enhance our mathematical proficiency.