Solving Radical Equations An In-Depth Guide To Finding X

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In the realm of mathematics, solving equations is a fundamental skill. Among the various types of equations, radical equations, which involve square roots or other radicals, often pose a unique challenge. This article delves into the step-by-step process of solving a specific radical equation and extracting the value of x. We will explore the techniques involved and provide a comprehensive understanding of how to tackle such problems.

Understanding the Equation

At the heart of our exploration lies the equation x = √((5x/2) + (1/6)) and 2x - 5 = 2. This equation presents a square root term, which necessitates a methodical approach to isolate and eliminate the radical. Before embarking on the solution, it's essential to grasp the nature of radical equations and the strategies employed to solve them.

The Essence of Radical Equations

Radical equations are algebraic expressions where the variable is ensnared within a radical symbol, such as a square root, cube root, or higher-order root. The presence of radicals introduces an added layer of complexity, as we need to carefully manipulate the equation to undo the radical and extract the variable. The key to solving radical equations lies in isolating the radical term and then raising both sides of the equation to a power that matches the index of the radical.

Strategies for Solving Radical Equations

When confronted with a radical equation, the following steps generally guide the solution process:

  1. Isolate the Radical: The primary objective is to isolate the radical term on one side of the equation. This involves performing algebraic operations, such as addition, subtraction, multiplication, or division, to move all other terms to the opposite side.
  2. Eliminate the Radical: Once the radical is isolated, we eliminate it by raising both sides of the equation to a power equal to the index of the radical. For a square root, we square both sides; for a cube root, we cube both sides, and so on. This step effectively undoes the radical, allowing us to work with a more familiar algebraic expression.
  3. Solve the Resulting Equation: After eliminating the radical, we are left with a simpler equation, typically a linear or quadratic equation. We then employ standard algebraic techniques to solve for the variable.
  4. Check for Extraneous Solutions: A crucial step in solving radical equations is to check for extraneous solutions. These are solutions that arise during the solving process but do not satisfy the original equation. Extraneous solutions often occur when squaring both sides of an equation, as this operation can introduce solutions that are not valid in the original context.

Solving the Given Equation

Now, let's apply these strategies to solve the equation x = √((5x/2) + (1/6)) and 2x - 5 = 2. Our goal is to find the value(s) of x that satisfy this equation.

Step 1 Isolate x from 2x - 5 = 2

First, we focus on the equation 2x - 5 = 2. To isolate x, we add 5 to both sides of the equation:

2x - 5 + 5 = 2 + 5

This simplifies to:

2x = 7

Next, we divide both sides by 2 to solve for x:

2x / 2 = 7 / 2

Therefore,

x = 7 / 2

Step 2 Substitute x = 7/2 into x = √((5x/2) + (1/6))

Now we substitute x = 7/2 into the original equation:

7/2 = √((5 * (7/2) / 2) + (1/6))

Simplify the expression inside the square root:

7/2 = √((35/4) + (1/6))

Find a common denominator for the fractions inside the square root, which is 12:

7/2 = √((105/12) + (2/12))

7/2 = √(107/12)

Step 3 Verify the Solution

To verify if x = 7/2 is a valid solution, we need to check if both sides of the equation are equal:

7/2 ≈ 3.5

√(107/12) ≈ √(8.9167) ≈ 2.986

Since 3.5 ≠ 2.986, the solution x = 7/2 does not satisfy the original equation

Step 4 Solve x = √((5x/2) + (1/6))

Square both sides of the equation to eliminate the square root:

x^2 = (5x/2) + (1/6)

Multiply both sides by 6 to eliminate the fractions:

6x^2 = 15x + 1

Rearrange the equation into a quadratic equation:

6x^2 - 15x - 1 = 0

Step 5 Apply the Quadratic Formula

Use the quadratic formula to solve for x:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 6, b = -15, and c = -1:

x = (15 ± √((-15)^2 - 4 * 6 * (-1))) / (2 * 6)

x = (15 ± √(225 + 24)) / 12

x = (15 ± √249) / 12

Step 6 Find the Two Possible Values of x

So, we have two possible solutions:

x = (15 + √249) / 12

x = (15 - √249) / 12

Step 7 Approximate the Values

Approximate the square root of 249:

√249 ≈ 15.78

Substitute this value back into the solutions:

x ≈ (15 + 15.78) / 12 ≈ 30.78 / 12 ≈ 2.565

x ≈ (15 - 15.78) / 12 ≈ -0.78 / 12 ≈ -0.065

Step 8 Check for Extraneous Solutions

We need to check both solutions in the original equation x = √((5x/2) + (1/6)).

Check x ≈ 2.565

2. 565 ≈ √((5 * 2.565 / 2) + (1/6))

2. 565 ≈ √(6.4125 + 0.1667)

2. 565 ≈ √6.5792

2. 565 ≈ 2.565

This solution is valid.

Check x ≈ -0.065

-0. 065 ≈ √((5 * -0.065 / 2) + (1/6))

Since the square root cannot be negative, this solution is extraneous.

Final Answer

Therefore, the value of x that satisfies the equation x = √((5x/2) + (1/6)) is approximately x ≈ 2.565.

Options Analysis

Let's analyze the given options based on our solution:

  • āĻ•) 3x: 3 * 2.565 ≈ 7.695
  • āĻ–) 1/(3x): 1 / (3 * 2.565) ≈ 0.1297
  • āĻ—) 2x: 2 * 2.565 ≈ 5.13
  • āĻ˜) 1/(2x): 1 / (2 * 2.565) ≈ 0.1949

None of these options directly give us the value of x, but they represent different expressions involving x. To determine which option is correct, we would need to substitute our solution back into the original equation and see which one holds true.

Conclusion

Solving radical equations requires a systematic approach that involves isolating the radical, eliminating it by raising both sides to an appropriate power, and solving the resulting equation. It is crucial to check for extraneous solutions, as these can arise during the solving process and do not represent valid solutions to the original equation. By following these steps and carefully verifying the solutions, we can confidently solve radical equations and extract the values of the variables involved.

This in-depth exploration of solving a radical equation provides a comprehensive understanding of the techniques involved and highlights the importance of methodical problem-solving in mathematics. Mastering these skills is essential for success in algebra and beyond.