Solving Quadratic Equations: A Step-by-Step Guide

by Alex Johnson 50 views

In this article, we will walk through solving the quadratic equation (v−4)2−54=0(v-4)^2 - 54 = 0 for the real number vv. We will simplify the solution as much as possible and provide a clear, step-by-step explanation. If there's more than one solution, we'll separate them with commas. Let's dive in!

Understanding the Problem

Our main task is to solve the equation

(v−4)2−54=0(v-4)^2 - 54 = 0

for the real number v. This equation is a quadratic equation, although it might not immediately look like the standard form ax2+bx+c=0ax^2 + bx + c = 0. Our strategy will involve isolating the squared term, then taking the square root of both sides. This approach allows us to find the values of v that satisfy the equation. The key to solving this equation efficiently lies in recognizing its structure and applying algebraic manipulations carefully. It's also important to remember that when taking the square root, we must consider both positive and negative roots to find all possible solutions for v. By following these steps, we can simplify the equation and determine the values of v that make the equation true. Understanding the problem thoroughly is the first step towards a successful solution. Remember, quadratic equations often have two solutions, one solution, or no real solutions, depending on the discriminant.

Step-by-Step Solution

First, let's isolate the squared term. We start with the equation:

(v−4)2−54=0(v-4)^2 - 54 = 0

Add 54 to both sides:

(v−4)2=54(v-4)^2 = 54

Now, take the square root of both sides of the equation:

v−4=±54v-4 = \pm\sqrt{54}

Simplify the square root. Since 54=9×654 = 9 \times 6, we have:

v−4=±9×6=±36v-4 = \pm\sqrt{9 \times 6} = \pm 3\sqrt{6}

Next, isolate v by adding 4 to both sides:

v=4±36v = 4 \pm 3\sqrt{6}

Thus, we have two solutions:

v1=4+36v_1 = 4 + 3\sqrt{6}

v2=4−36v_2 = 4 - 3\sqrt{6}

So, the solutions are 4+364 + 3\sqrt{6} and 4−364 - 3\sqrt{6}. Let's recap the steps to ensure clarity. We began by isolating the squared term, which involved adding 54 to both sides of the original equation. This gave us (v−4)2=54(v-4)^2 = 54. Next, we took the square root of both sides, remembering to include both the positive and negative roots, resulting in v−4=±54v-4 = \pm\sqrt{54}. We then simplified the square root of 54 to 363\sqrt{6}, leading to v−4=±36v-4 = \pm 3\sqrt{6}. Finally, we isolated v by adding 4 to both sides, yielding the two solutions v=4+36v = 4 + 3\sqrt{6} and v=4−36v = 4 - 3\sqrt{6}. These solutions represent the values of v that satisfy the original equation. The process involved basic algebraic manipulation and simplification of radicals, common techniques when solving quadratic equations.

Verification

To verify our solutions, we can plug them back into the original equation (v−4)2−54=0(v-4)^2 - 54 = 0. Let's start with v1=4+36v_1 = 4 + 3\sqrt{6}:

((4+36)−4)2−54=(36)2−54=9×6−54=54−54=0((4 + 3\sqrt{6}) - 4)^2 - 54 = (3\sqrt{6})^2 - 54 = 9 \times 6 - 54 = 54 - 54 = 0

Now, let's verify v2=4−36v_2 = 4 - 3\sqrt{6}:

((4−36)−4)2−54=(−36)2−54=9×6−54=54−54=0((4 - 3\sqrt{6}) - 4)^2 - 54 = (-3\sqrt{6})^2 - 54 = 9 \times 6 - 54 = 54 - 54 = 0

Both solutions satisfy the original equation. Therefore, our solutions are correct. Verification is an important step in solving any equation, as it helps confirm that the solutions obtained are indeed valid. By substituting the solutions back into the original equation, we can check if the equation holds true. In this case, both v1=4+36v_1 = 4 + 3\sqrt{6} and v2=4−36v_2 = 4 - 3\sqrt{6} resulted in the equation being satisfied, confirming their correctness. This process not only validates the solutions but also provides confidence in the algebraic manipulations performed during the solution process. Furthermore, it reinforces the understanding of the relationship between the solutions and the original equation.

Final Answer

The solutions to the equation (v−4)2−54=0(v-4)^2 - 54 = 0 are:

v=4+36,4−36v = 4 + 3\sqrt{6}, 4 - 3\sqrt{6}

Therefore, the final answer is 4+36,4−364 + 3\sqrt{6}, 4 - 3\sqrt{6}. This concludes our step-by-step solution. We've successfully found the values of v that satisfy the given equation and have verified the results to ensure accuracy. Solving quadratic equations like this one involves a combination of algebraic techniques, simplification, and careful attention to detail. By following these steps, you can confidently tackle similar problems in the future. Remember to always verify your solutions to ensure their validity and to reinforce your understanding of the underlying concepts. This systematic approach not only helps in solving mathematical problems but also enhances problem-solving skills in general. The ability to break down complex problems into smaller, manageable steps is a valuable skill that can be applied in various aspects of life.

For further learning on quadratic equations, you can visit Khan Academy's Quadratic Equations.