Solving Algebraic Equations, Perfect Squares, And Prime Factorization Problems

Let's dive into solving the algebraic equation 3(x+2)^2 - 124 = 200 - (x+2)^2 and unravel the mystery of finding the value of 'x'. This equation might seem daunting at first glance, but with a systematic approach, we can simplify it and arrive at the solution. The key to success here lies in recognizing the common term (x+2)^2 and treating it as a single variable. By doing so, we can significantly reduce the complexity of the equation and make it easier to manipulate.

First, let's consolidate the terms involving (x+2)^2. We can achieve this by adding (x+2)^2 to both sides of the equation. This gives us: 3(x+2)^2 + (x+2)^2 - 124 = 200. Notice how we've grouped the terms with the squared expression together. This is a crucial step in simplifying the equation and making it more manageable.

Next, we can combine the like terms on the left side. Since we have 3 instances of (x+2)^2 and we're adding another instance of it, we get a total of 4 instances. So, the equation becomes: 4(x+2)^2 - 124 = 200. This step is a simple yet effective way to further reduce the complexity of the equation.

Now, let's isolate the term with the squared expression. We can do this by adding 124 to both sides of the equation: 4(x+2)^2 = 200 + 124. This gives us 4(x+2)^2 = 324. We're getting closer to isolating the variable 'x'.

To further isolate (x+2)^2, we divide both sides of the equation by 4: (x+2)^2 = 324 / 4. This simplifies to (x+2)^2 = 81. Now we have a much simpler equation to work with.

At this point, we need to take the square root of both sides of the equation. Remember that when taking the square root, we need to consider both the positive and negative roots: √(x+2)^2 = ±√81. This gives us x+2 = ±9. It's essential to consider both positive and negative roots to ensure we find all possible solutions for 'x'.

Now we have two separate equations to solve: x+2 = 9 and x+2 = -9. Let's solve each one individually.

For the first equation, x+2 = 9, we subtract 2 from both sides to get x = 9 - 2, which simplifies to x = 7. This is one possible solution for 'x'.

For the second equation, x+2 = -9, we subtract 2 from both sides to get x = -9 - 2, which simplifies to x = -11. This is another possible solution for 'x'.

Therefore, the solutions to the equation are x = 7 and x = -11. Looking at the options provided (1) 5, 2) 7, 3) 3, 4) 4), we can see that option 2, which is 7, is one of the solutions we found. It is very important to double-check your work to ensure accuracy, especially in mathematical problems. Minor errors in calculation can lead to significant variations in the final result.

The next challenge is to identify which of the given numbers is not a perfect square: 1) 16384, 2) 23857, 3) 18496, 4) 11025. To determine whether a number is a perfect square, we need to understand what a perfect square is. A perfect square is an integer that can be expressed as the square of another integer. In simpler terms, it's a number you get when you multiply an integer by itself. For instance, 9 is a perfect square because it's 3 multiplied by 3 (3^2 = 9).

One of the most straightforward methods to check for perfect squares is by finding the square root of the number. If the square root is an integer, then the original number is a perfect square. However, calculating square roots, especially for large numbers, can be time-consuming. Therefore, it's often helpful to look for patterns and properties that can quickly rule out certain numbers.

Before diving into calculations, let's consider some key characteristics of perfect squares. The last digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. This is because when you square any integer, the last digit of the result is determined solely by the last digit of the original integer. For example, any number ending in 2, when squared, will end in 4 (2^2 = 4, 12^2 = 144, 22^2 = 484, etc.). Similarly, any number ending in 3, when squared, will end in 9. This observation helps us eliminate some options quickly.

Now let's apply this knowledge to the given options.

1. 16384: The last digit is 4, which is a possible last digit for a perfect square. To confirm, we can try finding its square root. The square root of 16384 is 128, which is an integer. Therefore, 16384 is a perfect square.

2. 23857: The last digit is 7. Since 7 is not among the possible last digits for a perfect square (0, 1, 4, 5, 6, or 9), we can confidently conclude that 23857 is not a perfect square. This eliminates the need for any further calculations.

3. 18496: The last digit is 6, which is a possible last digit for a perfect square. We can find its square root to confirm. The square root of 18496 is 136, which is an integer. So, 18496 is a perfect square.

4. 11025: The last digit is 5, which is a possible last digit for a perfect square. To verify, we find its square root. The square root of 11025 is 105, which is an integer. Thus, 11025 is a perfect square.

Based on our analysis, 23857 is the only number that is not a perfect square because its last digit is 7. This highlights the importance of understanding the properties of perfect squares as a tool for quickly identifying non-perfect squares.

Our final challenge involves finding the smallest number by which 6480 should be divided so that the quotient is a perfect square. This problem delves into the concept of prime factorization and how it relates to perfect squares. To solve this, we need to break down 6480 into its prime factors and identify the factors that need to be adjusted to make the number a perfect square.

Firstly, let's find the prime factorization of 6480. We can do this by repeatedly dividing the number by its smallest prime factor until we are left with 1. The prime factorization of 6480 is: 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 5, which can be written as 2^4 x 3^4 x 5^1. This step is crucial because it reveals the composition of the number in terms of its prime factors.

Now, let's understand what makes a number a perfect square in terms of its prime factors. A number is a perfect square if all the exponents in its prime factorization are even. This is because the square root of such a number will have integer exponents in its prime factorization. For example, the square root of 2^4 is 2^2, and the square root of 3^2 is 3^1.

Looking at the prime factorization of 6480 (2^4 x 3^4 x 5^1), we can see that the exponents of 2 and 3 are both 4, which are even numbers. However, the exponent of 5 is 1, which is an odd number. This means that 6480 is not a perfect square as it stands.

To make the quotient a perfect square, we need to divide 6480 by a number that will result in all even exponents in the prime factorization of the quotient. In this case, the only factor with an odd exponent is 5^1. Therefore, we need to divide 6480 by 5 to reduce the exponent of 5 to 0, effectively removing it from the prime factorization.

Dividing 6480 by 5 gives us 1296. Let's find the prime factorization of 1296 to confirm that it is a perfect square. The prime factorization of 1296 is 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3, which can be written as 2^4 x 3^4. Now we see that all exponents are even, confirming that 1296 is a perfect square (36^2 = 1296).

Therefore, the smallest number by which 6480 should be divided so that the quotient is a perfect square is 5. This demonstrates how understanding prime factorization and the properties of perfect squares can help us solve such problems efficiently.

In conclusion, by systematically analyzing the prime factors of a number, we can determine the smallest divisor required to obtain a perfect square quotient. This approach not only solves the problem but also provides a deeper understanding of the mathematical concepts involved.

Mathematics, often perceived as a challenging subject, is in reality a fascinating field filled with logic, patterns, and problem-solving opportunities. The key to mastering mathematics lies in understanding the fundamental concepts, practicing regularly, and developing a strategic approach to solving problems. In this article, we've explored three distinct mathematical problems, each requiring a unique set of skills and knowledge. Let's recap the key strategies and concepts we've utilized and how they can be applied to a broader range of mathematical challenges.

Firstly, let's revisit the algebraic equation 3(x+2)^2 - 124 = 200 - (x+2)^2. The core strategy here was simplification. By recognizing the common term (x+2)^2, we treated it as a single variable, which significantly reduced the complexity of the equation. This technique of substitution is a powerful tool in algebra and can be applied to various types of equations, including quadratics, polynomials, and even trigonometric equations. The key is to identify repeating expressions and temporarily replace them with a single variable to make the equation more manageable.

Furthermore, we emphasized the importance of considering both positive and negative roots when taking the square root of both sides of an equation. This is a crucial step often overlooked, leading to incomplete solutions. Remember, the square root of a number has two possible values: a positive and a negative one. In our example, this led to two distinct solutions for 'x', highlighting the necessity of this step.

Next, we tackled the problem of identifying non-perfect squares. This involved understanding the definition of a perfect square and recognizing patterns associated with them. We discovered that the last digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. This simple rule allowed us to quickly eliminate options without performing complex calculations. This exemplifies the importance of pattern recognition in mathematics. By observing patterns and properties, we can develop shortcuts and efficient problem-solving techniques.

In addition, we discussed the method of finding the square root to confirm whether a number is a perfect square. While this method is straightforward, it can be time-consuming for large numbers. Therefore, combining it with pattern recognition and properties of perfect squares is the most effective approach. Mastering these techniques can significantly improve your speed and accuracy in solving problems related to perfect squares.

Finally, we addressed the problem of finding the smallest divisor to make a quotient a perfect square. This problem required us to delve into prime factorization. By breaking down the number into its prime factors, we could easily identify the factors with odd exponents. Remember, a number is a perfect square if all the exponents in its prime factorization are even. Therefore, we needed to divide the number by a factor that would make all exponents even.

This problem underscores the significance of prime factorization in number theory. It's a fundamental concept with applications in various areas of mathematics, including finding the greatest common divisor (GCD), the least common multiple (LCM), and simplifying fractions. A solid understanding of prime factorization is crucial for success in many mathematical contexts.

In conclusion, solving mathematical problems effectively involves a combination of understanding fundamental concepts, employing strategic techniques, and practicing regularly. Whether it's simplifying algebraic equations, recognizing patterns in perfect squares, or utilizing prime factorization, each problem provides an opportunity to enhance your mathematical skills and problem-solving abilities. Embrace the challenges, learn from your mistakes, and celebrate your successes. With dedication and the right approach, you can unlock the beauty and power of mathematics.

Algebraic equation, Perfect square, Prime factorization, Mathematical problem-solving, Simplify equation, Square root, Non-perfect squares, Divisor, Quotient, Mathematical concepts, Problem-solving strategies, Practice mathematics, Mathematical skills, Pattern recognition, Number theory.