Solving Quadratic Equations Using The Square Root Property For U^2 - 40u + 400 = 16

Understanding the Square Root Property

The square root property is a powerful tool for solving quadratic equations, especially those that can be easily expressed in the form $(x + a)^2 = b$ or $(x - a)^2 = b$. This method simplifies the process of finding the solutions (or roots) of a quadratic equation by directly addressing the squared term. To effectively utilize the square root property, it is crucial to first isolate the squared expression on one side of the equation and then take the square root of both sides. When taking the square root, remember to consider both the positive and negative roots, as both will satisfy the original equation. This is because squaring either a positive or a negative number yields a positive result. The square root property provides a straightforward approach to solving certain quadratic equations, avoiding the more complex procedures like factoring or using the quadratic formula in suitable cases. By understanding and applying this property, you can efficiently determine the solutions to various quadratic equations, enhancing your problem-solving skills in algebra and beyond. The square root property is particularly useful when the quadratic equation is already in, or can be easily manipulated into, the form where a perfect square is isolated on one side, making it a valuable technique in your mathematical toolkit. When dealing with equations involving square roots, it's also essential to check your solutions to ensure they are valid and do not introduce extraneous roots, which can sometimes occur when dealing with radical equations. This step is crucial for ensuring the accuracy of your solutions and maintaining a solid understanding of the underlying principles of equation solving. The square root property not only simplifies the process but also deepens your comprehension of the relationship between squares and square roots in algebraic equations.

Applying the Square Root Property to the Equation $u^2 - 40u + 400 = 16$

To solve the quadratic equation $u^2 - 40u + 400 = 16$ using the square root property, our initial step involves recognizing that the left-hand side of the equation is a perfect square trinomial. This is a crucial observation because it allows us to rewrite the equation in a form that is suitable for applying the square root property. The expression $u^2 - 40u + 400$ can be factored as $(u - 20)^2$. This transformation is key to simplifying the equation and making it solvable using the square root property. Once we've identified and factored the perfect square trinomial, the equation becomes $(u - 20)^2 = 16$. Now, we have the equation in the desired form, where a squared expression is isolated on one side. The next step is to take the square root of both sides of the equation. Remember that when we take the square root of a number, we must consider both the positive and negative roots. This is because both the positive and negative values, when squared, will result in the same positive number. Therefore, taking the square root of both sides of $(u - 20)^2 = 16$ gives us $u - 20 = ±\sqrt{16}$. Since the square root of 16 is 4, we have $u - 20 = ±4$. This equation represents two separate equations: $u - 20 = 4$ and $u - 20 = -4$. Solving these two equations will give us the two possible values for $u$ that satisfy the original quadratic equation. By carefully applying the square root property and remembering to consider both positive and negative roots, we can efficiently solve quadratic equations of this form. This method not only provides the solutions but also reinforces the understanding of perfect square trinomials and their relationship to the square root property.

Solving for u

Having transformed the original equation into $(u - 20)^2 = 16$ and taken the square root of both sides to obtain $u - 20 = ±4$, we now have two simple linear equations to solve for $u$. These equations are $u - 20 = 4$ and $u - 20 = -4$. To solve the first equation, $u - 20 = 4$, we isolate $u$ by adding 20 to both sides of the equation. This gives us $u = 4 + 20$, which simplifies to $u = 24$. This is one of the solutions to the original quadratic equation. Next, we solve the second equation, $u - 20 = -4$. Again, we isolate $u$ by adding 20 to both sides of the equation. This gives us $u = -4 + 20$, which simplifies to $u = 16$. This is the second solution to the original quadratic equation. Therefore, the solutions to the quadratic equation $u^2 - 40u + 400 = 16$ are $u = 24$ and $u = 16$. It's essential to verify these solutions by substituting them back into the original equation to ensure they satisfy the equation. This step helps to avoid errors and confirms the accuracy of our solutions. By solving the two linear equations derived from the square root property, we have successfully found the values of $u$ that make the original quadratic equation true. This process demonstrates the power and efficiency of the square root property in solving certain types of quadratic equations. The ability to break down the problem into simpler steps and solve for the variable is a fundamental skill in algebra and crucial for more advanced mathematical concepts.

Verifying the Solutions

To ensure the accuracy of our solutions, it is crucial to verify them by substituting each value of $u$ back into the original equation, $u^2 - 40u + 400 = 16$. This process confirms that the solutions we obtained, $u = 24$ and $u = 16$, indeed satisfy the equation. Let's start by verifying the solution $u = 24$. Substituting $u = 24$ into the equation, we get $(24)^2 - 40(24) + 400$. Calculating this, we have $576 - 960 + 400$, which simplifies to $16$. Since this result matches the right-hand side of the original equation, $u = 24$ is indeed a valid solution. Now, let's verify the solution $u = 16$. Substituting $u = 16$ into the equation, we get $(16)^2 - 40(16) + 400$. Calculating this, we have $256 - 640 + 400$, which also simplifies to $16$. This result also matches the right-hand side of the original equation, confirming that $u = 16$ is a valid solution as well. The verification process is an essential step in solving equations because it helps identify any potential errors made during the solution process. It ensures that the values obtained for the variable actually make the equation true. By verifying our solutions, we can have confidence in the accuracy of our work and the correctness of our answers. This step is particularly important when dealing with more complex equations or when using methods that can sometimes introduce extraneous solutions. The habit of verifying solutions is a hallmark of a careful and thorough mathematical problem-solver.

Final Answer

After successfully applying the square root property to the quadratic equation $u^2 - 40u + 400 = 16$ and verifying our solutions, we have determined that the values of $u$ that satisfy the equation are $24$ and $16$. Therefore, the final answer is $16, 24$. These values represent the roots of the quadratic equation, meaning they are the points where the corresponding parabola intersects the x-axis. The process of solving this equation involved several key steps, starting with recognizing the perfect square trinomial on the left-hand side, factoring it, and then applying the square root property. This method allowed us to transform the quadratic equation into two simpler linear equations, which were then easily solved for $u$. The verification step further ensured the accuracy of our solutions, providing confidence in the final answer. Understanding and applying the square root property is a valuable skill in algebra, particularly for solving quadratic equations that can be expressed in the form $(x - a)^2 = b$ or $(x + a)^2 = b$. This method not only simplifies the solution process but also enhances our understanding of the relationship between squares and square roots. The final answer, $16, 24$, represents the complete solution set for the given quadratic equation, providing a clear and concise answer to the problem. This comprehensive approach to solving quadratic equations demonstrates a strong foundation in algebraic principles and problem-solving techniques.