Solving The Inequality \( \frac{\sqrt{a+4}}{1-a} < 1 \) A Comprehensive Guide

Introduction

In this comprehensive article, we will delve into the intricate process of solving the inequality ${ \frac{\sqrt{a+4}}{1-a} < 1 }$. This problem falls under the domain of mathematical inequalities, requiring a meticulous approach to determine the range of values for 'a' that satisfy the given condition. We will explore the various steps involved, from identifying the domain of the expression to handling potential sign changes and ensuring the validity of our solutions. The correct option, as we will demonstrate, is (a) ${ \left[ -4, \frac{3-\sqrt{21}}{2} \right] \cup (1,\infty) }$, and this article will serve as a detailed guide to understanding why.

Understanding the Problem

The core of this problem lies in understanding how to handle inequalities involving square roots and rational expressions. We must consider the restrictions imposed by the square root (the radicand must be non-negative) and the denominator (which cannot be zero). Furthermore, the sign of ${ 1-a }$ plays a crucial role in how we manipulate the inequality. The goal is to isolate 'a' while ensuring that all operations performed maintain the validity of the inequality.

Domain of the Expression

Before we begin manipulating the inequality, we must first establish the domain of the expression. The square root term ${ \sqrt{a+4} }$ requires that ${ a+4 \ge 0 }$, which implies ${ a \ge -4 }$. Additionally, the denominator ${ 1-a }$ cannot be zero, so ${ a \ne 1 }$. Therefore, the domain of the expression is ${ a \in [-4, 1) \cup (1, \infty) }$.

Analyzing the Inequality

To solve the inequality ${ \frac{\sqrt{a+4}}{1-a} < 1 }$, we need to consider two cases based on the sign of ${ 1-a }$:

Case 1: ${ 1-a > 0 }$ or ${ a < 1 }$

When ${ 1-a }$ is positive, we can multiply both sides of the inequality by ${ 1-a }$ without changing the direction of the inequality:

${ \sqrt{a+4} < 1-a }$

Since both sides are non-negative (because ${ \sqrt{a+4} }$ is always non-negative and we are considering the case where ${ 1-a > 0 }$), we can square both sides:

${ a+4 < (1-a)^2 }$

${ a+4 < 1 - 2a + a^2 }$

${ 0 < a^2 - 3a - 3 }$

Now, we need to find the roots of the quadratic equation ${ a^2 - 3a - 3 = 0 }$. Using the quadratic formula, we get:

${ a = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-3)}}{2(1)} }$

${ a = \frac{3 \pm \sqrt{9 + 12}}{2} }$

${ a = \frac{3 \pm \sqrt{21}}{2} }$

So, the roots are ${ a_1 = \frac{3 - \sqrt{21}}{2} }$ and ${ a_2 = \frac{3 + \sqrt{21}}{2} }$. Since the quadratic expression ${ a^2 - 3a - 3 }$ is positive when 'a' is outside the interval between the roots, we have:

${ a < \frac{3 - \sqrt{21}}{2} \text{ or } a > \frac{3 + \sqrt{21}}{2} }$

However, we are in the case where ${ a < 1 }$. Also, we must consider the domain restriction ${ a \ge -4 }$. Thus, we have:

${ -4 \le a < \frac{3 - \sqrt{21}}{2} }$

which gives us the interval ${ \left[ -4, \frac{3-\sqrt{21}}{2} \right] }$.

Case 2: ${ 1-a < 0 }$ or ${ a > 1 }$

When ${ 1-a }$ is negative, multiplying both sides of the inequality by ${ 1-a }$ reverses the direction of the inequality:

${ \sqrt{a+4} > 1-a }$

Since ${ \sqrt{a+4} }$ is non-negative and ${ 1-a }$ is negative, this inequality is always true. However, we must consider the domain restriction ${ a > 1 }$. Thus, the solution in this case is ${ (1, \infty) }$.

Combining the Solutions

Combining the solutions from both cases, we get:

${ a \in \left[ -4, \frac{3-\sqrt{21}}{2} \right] \cup (1,\infty) }$

This corresponds to option (a).

Detailed Solution Walkthrough

Let's break down the solution step-by-step to ensure clarity.

1. Identify the Domain: The expression ${ \frac{\sqrt{a+4}}{1-a} }$ is defined when ${ a+4 \ge 0 }$ and ${ 1-a \ne 0 }$. This gives us ${ a \ge -4 }$ and ${ a \ne 1 }$. Thus, the domain is ${ [-4, 1) \cup (1, \infty) }$.
2. Consider Cases Based on the Sign of ${ 1-a }$: We need to consider two cases:
   • Case 1: ${ 1-a > 0 }$ (i.e., ${ a < 1 }$)
   • Case 2: ${ 1-a < 0 }$ (i.e., ${ a > 1 }$)
3. Case 1: ${ 1-a > 0 }$ (or ${ a < 1 }$): Multiply both sides of the inequality by ${ 1-a }$ (without changing the inequality sign) to get: ${ \sqrt{a+4} < 1-a }$ Square both sides (since both sides are non-negative): ${ a+4 < (1-a)^2 }$ ${ a+4 < 1 - 2a + a^2 }$ Rearrange to get a quadratic inequality: ${ 0 < a^2 - 3a - 3 }$ Find the roots of the quadratic equation ${ a^2 - 3a - 3 = 0 }$ using the quadratic formula: ${ a = \frac{3 \pm \sqrt{21}}{2} }$ The roots are ${ a_1 = \frac{3 - \sqrt{21}}{2} \approx -0.79 }$ and ${ a_2 = \frac{3 + \sqrt{21}}{2} \approx 3.79 }$. The inequality ${ a^2 - 3a - 3 > 0 }$ is satisfied when ${ a < \frac{3 - \sqrt{21}}{2} }$ or ${ a > \frac{3 + \sqrt{21}}{2} }$. Considering ${ a < 1 }$ and ${ a \ge -4 }$, the solution for this case is ${ \left[ -4, \frac{3 - \sqrt{21}}{2} \right] }$.
4. Case 2: ${ 1-a < 0 }$ (or ${ a > 1 }$): Multiply both sides of the inequality by ${ 1-a }$ (and reverse the inequality sign): ${ \sqrt{a+4} > 1-a }$ Since ${ \sqrt{a+4} }$ is non-negative and ${ 1-a }$ is negative, this inequality is always true for ${ a > 1 }$. Thus, the solution for this case is ${ (1, \infty) }$.
5. Combine the Solutions: Combining the solutions from both cases, we get: ${ a \in \left[ -4, \frac{3 - \sqrt{21}}{2} \right] \cup (1, \infty) }$

Common Pitfalls and How to Avoid Them

Solving inequalities, especially those involving square roots and rational expressions, can be tricky. Here are some common pitfalls to watch out for:

• Forgetting the Domain: Always start by determining the domain of the expression. Square roots require non-negative radicands, and denominators cannot be zero. Ignoring these restrictions can lead to extraneous solutions.
• Not Considering Sign Changes: When multiplying or dividing an inequality by an expression, the sign of the expression matters. If the expression is negative, the direction of the inequality must be reversed.
• Squaring Both Sides Without Verification: Squaring both sides of an inequality is only valid if both sides are non-negative. If one side is negative, squaring can introduce extraneous solutions. Always check your solutions in the original inequality.
• Incorrectly Solving Quadratic Inequalities: Make sure you correctly find the roots of the quadratic and understand how the sign of the quadratic expression changes in different intervals.

Conclusion

In conclusion, the solution to the inequality ${ \frac{\sqrt{a+4}}{1-a} < 1 }$ is ${ a \in \left[ -4, \frac{3-\sqrt{21}}{2} \right] \cup (1,\infty) }$, which corresponds to option (a). This solution is obtained by carefully considering the domain of the expression, analyzing the inequality in different cases based on the sign of ${ 1-a }$, and avoiding common pitfalls such as forgetting the domain or incorrectly squaring both sides of the inequality. By following a systematic approach and paying close attention to detail, we can successfully solve this type of mathematical problem.

This article provides a comprehensive guide to solving this particular inequality, demonstrating the necessary steps and reasoning involved. The detailed solution walkthrough and the discussion of common pitfalls aim to enhance understanding and problem-solving skills in the realm of mathematical inequalities. The key takeaway is the importance of a methodical approach, considering all constraints and potential sign changes, to arrive at the correct solution. Remember, understanding the underlying principles is just as crucial as the solution itself, if not more so.

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