Unit 6 Radical Functions Homework 1 Answer Key

Unit 6 Radical Functions Homework 1: A Comprehensive Guide to Solving Radical Equations and Inequalities

This comprehensive guide delves into the solutions for a typical Unit 6 Radical Functions Homework 1 assignment. We'll cover various problem types involving radical equations and inequalities, providing detailed explanations and strategies to help you master these concepts. Remember, understanding the underlying principles is key to success in algebra. This isn't just about getting the answers; it's about building a strong foundation in working with radical expressions.

Understanding Radical Functions and Their Properties

Before diving into the homework problems, let's refresh our understanding of radical functions. A radical function is a function that contains a radical expression, typically a square root, but can also include cube roots, fourth roots, and so on. The general form of a radical function is:

f(x) = √(g(x))

where g(x) is some function inside the radical. It's crucial to remember the domain restrictions of radical functions. For example, with square roots, the expression inside the radical (the radicand) must be non-negative: g(x) ≥ 0. This restriction significantly impacts the function's domain and range. Higher-order roots (cube roots, etc.) have less restrictive domains.

Key Properties to Remember:

• Squaring both sides: A common technique when solving radical equations is to square both sides to eliminate the radical. However, this step can introduce extraneous solutions, which are solutions that satisfy the squared equation but not the original radical equation. Always check your solutions in the original equation.

• Isolating the radical: Before squaring both sides, it's generally best to isolate the radical term on one side of the equation.

• Domain considerations: Always check if your solutions are within the domain of the original radical function.

Solving Radical Equations: A Step-by-Step Approach

Let's work through some example problems that could appear in your Unit 6 Radical Functions Homework 1 assignment.

Example 1: Solving a Simple Square Root Equation

Solve for x: √(x + 2) = 3

Solution:

1. Square both sides: (√(x + 2))² = 3² This simplifies to x + 2 = 9

2. Solve for x: x = 9 - 2 = 7

3. Check for extraneous solutions: Substitute x = 7 back into the original equation: √(7 + 2) = √9 = 3. The solution is valid.

Therefore, x = 7

Example 2: Solving a Radical Equation with Multiple Radicals

Solve for x: √(x + 5) + √(x) = 5

Solution:

1. Isolate one radical: Subtract √x from both sides: √(x + 5) = 5 - √x

2. Square both sides: (√(x + 5))² = (5 - √x)² This gives: x + 5 = 25 - 10√x + x

3. Simplify and isolate the remaining radical: Subtract x from both sides: 5 = 25 - 10√x

4. Solve for the radical: -20 = -10√x => 2 = √x

5. Square both sides again: 4 = x

6. Check for extraneous solutions: Substitute x = 4 into the original equation: √(4 + 5) + √4 = √9 + 2 = 5. The solution is valid.

Therefore, x = 4

Example 3: Solving a Radical Equation with a Higher-Order Root

Solve for x: ³√(2x - 1) = 2

Solution:

1. Cube both sides: (³√(2x - 1))³ = 2³ This simplifies to 2x - 1 = 8

2. Solve for x: 2x = 9 => x = 9/2 = 4.5

3. Check for extraneous solutions: Substitute x = 4.5 into the original equation: ³√(2(4.5) - 1) = ³√(9 - 1) = ³√8 = 2. The solution is valid.

Therefore, x = 4.5

Solving Radical Inequalities: A Similar Approach with Added Considerations

Solving radical inequalities involves similar steps to solving radical equations, but with additional considerations for the inequality symbol. Remember to always consider the domain restrictions of the radical expression.

Example 4: Solving a Simple Radical Inequality

Solve for x: √(x - 1) < 3

Solution:

1. Square both sides: (√(x - 1))² < 3² This gives: x - 1 < 9

2. Solve for x: x < 10

3. Consider the domain: Since we have a square root, we must have x - 1 ≥ 0, which means x ≥ 1.

4. Combine the conditions: We need both x < 10 and x ≥ 1, so the solution is 1 ≤ x < 10.

Therefore, the solution is the interval [1, 10).

Example 5: Solving a More Complex Radical Inequality

Solve for x: √(2x + 1) ≥ x

Solution:

1. Square both sides: (√(2x + 1))² ≥ x² This gives: 2x + 1 ≥ x²

2. Rearrange into a quadratic inequality: x² - 2x - 1 ≤ 0

3. Solve the quadratic equation: x² - 2x - 1 = 0 Using the quadratic formula, we get x = (2 ± √8)/2 ≈ 2.414 or -0.414

4. Consider the parabola: Since the parabola opens upwards, the inequality x² - 2x - 1 ≤ 0 is satisfied between the roots.

5. Consider the domain: We need 2x + 1 ≥ 0, which implies x ≥ -1/2.

6. Combine the conditions: The solution is approximately -0.414 ≤ x ≤ 2.414, but constrained by x ≥ -1/2. Therefore, the final solution is approximately [-0.5, 2.414].

Therefore, the approximate solution is the interval [-0.5, 2.414].

Advanced Techniques and Problem Types

Unit 6 Radical Functions Homework 1 might include more advanced problems requiring techniques beyond simple squaring. These could involve:

• Rationalizing the denominator: If you encounter fractions with radicals in the denominator, you'll need to rationalize the denominator to simplify the expression.

• Substitution: In some cases, substitution can simplify a complex radical equation or inequality.

• Graphing: Graphing the function can help visualize the solution set, especially for inequalities.

Checking Your Work: The Importance of Verification

It is absolutely crucial to check your solutions in the original equation or inequality. This step helps identify extraneous solutions, ensuring accuracy. Don't skip this vital step!

This detailed guide offers a comprehensive approach to tackling the problems typically found in a Unit 6 Radical Functions Homework 1 assignment. Remember to practice consistently, focusing on understanding the underlying concepts and techniques. By mastering these, you'll build a strong foundation for more advanced algebraic topics. Good luck!