Solving For X In X^2 = 24 A Step-by-Step Guide

Understanding the Equation

In this mathematical exploration, we embark on a journey to solve for x in the equation ${ x^2 = 24 }$. This seemingly simple equation holds the key to unlocking fundamental concepts in algebra and number theory. The equation states that the square of a certain number, denoted by x, is equal to 24. Our objective is to find the values of x that satisfy this condition. Before diving into the solution, let's first delve into the significance of this equation and its implications. The equation ${ x^2 = 24 }$ is a quadratic equation, a type of polynomial equation where the highest power of the variable is 2. Quadratic equations are prevalent in various fields of mathematics, physics, and engineering. They often arise when modeling phenomena involving parabolic trajectories, such as the motion of projectiles or the shape of suspension cables. The solutions to a quadratic equation, also known as its roots, represent the points where the parabola intersects the x-axis. In the context of our equation, finding the values of x that satisfy ${ x^2 = 24 }$ is akin to determining the x-coordinates of the points where the parabola represented by the equation intersects the horizontal line y = 24. Furthermore, understanding the properties of quadratic equations allows us to develop techniques for solving more complex equations and mathematical problems. The process of solving ${ x^2 = 24 }$ involves algebraic manipulations and the application of fundamental mathematical principles. By carefully isolating x, we can unveil the values that satisfy the equation. This exploration not only provides a solution to a specific equation but also enhances our understanding of algebraic concepts and problem-solving strategies. Understanding the equation ${ x^2 = 24 }$ is crucial for grasping the broader context of mathematical problem-solving and its applications in various scientific and engineering disciplines. By dissecting the equation and employing appropriate mathematical techniques, we can decipher the values of x that satisfy the given condition.

Methods to Solve the Equation

To solve for x in the equation ${ x^2 = 24 }$, we can employ several methods, each offering a unique approach to unraveling the solution. One common method is to utilize the concept of square roots. Since the equation involves x squared, we can take the square root of both sides of the equation to isolate x. However, it's crucial to remember that when taking the square root, we must consider both the positive and negative roots. This stems from the fact that both a positive number and its negative counterpart, when squared, yield the same positive result. Another approach involves rearranging the equation into a standard quadratic form and then applying techniques such as factoring or the quadratic formula. While this method may seem more complex for a simple equation like ${ x^2 = 24 }$, it becomes invaluable when dealing with more intricate quadratic equations. The standard form of a quadratic equation is ${ ax^2 + bx + c = 0 }$, where a, b, and c are constants. By rearranging the equation ${ x^2 = 24 }$, we can rewrite it as ${ x^2 - 24 = 0 }$. In this form, we can attempt to factor the quadratic expression or utilize the quadratic formula to find the solutions. Furthermore, graphical methods can also be employed to approximate the solutions. By plotting the graph of the equation ${ y = x^2 }$ and the horizontal line ${ y = 24 }$, we can visually identify the points of intersection, which correspond to the solutions of the equation. This graphical approach provides a visual representation of the solutions and can be particularly useful for estimating solutions when analytical methods are challenging to apply. Moreover, numerical methods, such as iterative techniques, can be employed to approximate the solutions to any desired degree of accuracy. These methods involve making successive approximations until a solution is reached within a specified tolerance. While numerical methods may not provide exact solutions, they are valuable tools for solving equations that lack analytical solutions. In summary, multiple methods exist to solve for x in the equation ${ x^2 = 24 }$, each with its strengths and limitations. The choice of method depends on the specific equation and the desired level of accuracy.

Step-by-Step Solution

Let's embark on a step-by-step journey to solve for x in the equation ${ x^2 = 24 }$. Our approach will primarily involve utilizing the concept of square roots, a fundamental tool in algebra. First, we begin with the given equation: ${ x^2 = 24 }$. Our objective is to isolate x on one side of the equation. To achieve this, we take the square root of both sides of the equation. Remember, when taking the square root, we must consider both the positive and negative roots. Applying this principle, we have: ${ \sqrt{x^2} = \pm \sqrt{24} }$. The square root of ${ x^2 }$ is simply x, so we get: ${ x = \pm \sqrt{24} }$. Now, we need to simplify the square root of 24. We can express 24 as the product of its prime factors: ${ 24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3 }$. Using the properties of square roots, we can rewrite ${ \sqrt{24} }$ as: ${ \sqrt{24} = \sqrt{2^2 \times 2 \times 3} = \sqrt{2^2} \times \sqrt{2 \times 3} = 2\sqrt{6} }$. Therefore, our solution becomes: ${ x = \pm 2\sqrt{6} }$. This means that there are two possible values for x: ${ x = 2\sqrt{6} }$ and ${ x = -2\sqrt{6} }$. To obtain approximate decimal values for these solutions, we can use a calculator. The square root of 6 is approximately 2.449. Multiplying this by 2, we get approximately 4.899. Therefore, the approximate solutions are: ${ x \approx 4.899 }$ and ${ x \approx -4.899 }$. These values represent the points where the parabola represented by the equation ${ y = x^2 }$ intersects the horizontal line ${ y = 24 }$. In summary, by taking the square root of both sides of the equation and simplifying, we have successfully solved for x in the equation ${ x^2 = 24 }$. The solutions are ${ x = 2\sqrt{6} }$ and ${ x = -2\sqrt{6} }$, with approximate decimal values of 4.899 and -4.899, respectively.

Verification of the Solution

To ensure the accuracy of our solution, it's crucial to verify the values of x we obtained for the equation ${ x^2 = 24 }$. We found that the solutions are ${ x = 2\sqrt{6} }$ and ${ x = -2\sqrt{6} }$. To verify these solutions, we will substitute each value back into the original equation and check if it holds true. First, let's consider the positive solution, ${ x = 2\sqrt{6} }$. Substituting this value into the equation ${ x^2 = 24 }$, we get: ${ (2\sqrt{6})^2 = 24 }$. Squaring the expression on the left side, we have: ${ 2^2 \times (\sqrt{6})^2 = 4 \times 6 = 24 }$. Since this is equal to the right side of the equation, the solution ${ x = 2\sqrt{6} }$ is verified. Now, let's consider the negative solution, ${ x = -2\sqrt{6} }$. Substituting this value into the equation ${ x^2 = 24 }$, we get: ${ (-2\sqrt{6})^2 = 24 }$. Squaring the expression on the left side, we have: ${ (-2)^2 \times (\sqrt{6})^2 = 4 \times 6 = 24 }$. Again, this is equal to the right side of the equation, so the solution ${ x = -2\sqrt{6} }$ is also verified. Both the positive and negative solutions satisfy the original equation, confirming their validity. This verification process underscores the importance of checking solutions to ensure accuracy, particularly when dealing with equations involving square roots or other mathematical operations that can introduce extraneous solutions. In this case, we have demonstrated that both ${ x = 2\sqrt{6} }$ and ${ x = -2\sqrt{6} }$ are indeed valid solutions to the equation ${ x^2 = 24 }$. Furthermore, we can also verify the approximate decimal values we obtained earlier. We found that ${ x \approx 4.899 }$ and ${ x \approx -4.899 }$. Squaring these values, we get: ${ (4.899)^2 \approx 24.000201 }$ and ${ (-4.899)^2 \approx 24.000201 }$. These values are very close to 24, which further supports the accuracy of our solutions. In conclusion, through substitution and calculation, we have rigorously verified the solutions for x in the equation ${ x^2 = 24 }$, reinforcing our confidence in the correctness of our results.

Conclusion

In summary, we have successfully navigated the process of solving for x in the equation ${ x^2 = 24 }$. Our journey began with a thorough understanding of the equation, recognizing it as a quadratic equation with two potential solutions. We explored various methods for solving the equation, including utilizing square roots, rearranging into standard quadratic form, and employing graphical and numerical techniques. Ultimately, we opted for the method of taking the square root of both sides, which proved to be the most straightforward approach for this particular equation. Through careful algebraic manipulation and simplification, we arrived at the solutions ${ x = 2\sqrt{6} }$ and ${ x = -2\sqrt{6} }$. These solutions represent the values of x that, when squared, yield 24. Furthermore, we obtained approximate decimal values for these solutions, which provided a tangible sense of their magnitude. To ensure the accuracy of our results, we diligently verified the solutions by substituting them back into the original equation. This verification process confirmed that both ${ x = 2\sqrt{6} }$ and ${ x = -2\sqrt{6} }$ indeed satisfy the equation ${ x^2 = 24 }$. The process of solving for x in this equation not only demonstrates the application of fundamental algebraic principles but also highlights the importance of precision and accuracy in mathematical problem-solving. By understanding the underlying concepts and employing appropriate techniques, we can confidently tackle a wide range of mathematical challenges. Moreover, the skills and insights gained from solving this equation can be readily applied to more complex problems in mathematics, science, and engineering. The ability to manipulate equations, isolate variables, and verify solutions is a cornerstone of mathematical proficiency. In conclusion, our exploration of the equation ${ x^2 = 24 }$ has been a rewarding exercise in algebraic problem-solving. We have not only determined the solutions for x but also reinforced our understanding of fundamental mathematical concepts and techniques. The successful completion of this task underscores the power of mathematical reasoning and its ability to unravel the mysteries of the numerical world.