What Is The Solution Of $\frac{x^2-1}{x^2+5x+4} \leq 0$?A. $-4 \ \textless \ X \ \textless \ 1$ And $x \neq -1$B. $x \ \textless \ -4$ Or $x \geq 1$C. $x \ \textless \ -4$ Or $x \

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Introduction

In mathematics, solving inequalities is a crucial aspect of algebra and calculus. Inequalities can be solved using various methods, including factoring, graphing, and using the sign chart. In this article, we will focus on solving the inequality $\frac{x^2-1}{x^2+5x+4} \leq 0$. We will use the sign chart method to find the solution of this inequality.

Understanding the Inequality

The given inequality is $\frac{x^2-1}{x^2+5x+4} \leq 0$. To solve this inequality, we need to find the values of $x$ that make the expression $\frac{x^2-1}{x^2+5x+4}$ less than or equal to zero.

Factoring the Numerator and Denominator

To solve the inequality, we first need to factor the numerator and denominator. The numerator can be factored as $(x-1)(x+1)$, and the denominator can be factored as $(x+1)(x+4)$.

Creating a Sign Chart

Now that we have factored the numerator and denominator, we can create a sign chart to find the solution of the inequality. A sign chart is a table that shows the sign of the expression in different intervals.

Interval	$(x-1)$	$(x+1)$	$(x+4)$	$\frac{x^2-1}{x^2+5x+4}$
$(-\infty, -4)$	-	-	-	+
$(-4, -1)$	-	-	+	-
$(-1, 1)$	-	+	+	+
$(1, \infty)$	+	+	+	+

Analyzing the Sign Chart

From the sign chart, we can see that the expression $\frac{x^2-1}{x^2+5x+4}$ is less than or equal to zero in the intervals $(-4, -1)$ and $(1, \infty)$.

Finding the Solution

However, we need to exclude the values of $x$ that make the denominator zero. The denominator is zero when $x=-1$ and $x=-4$. Therefore, the solution of the inequality is $x \in (-4, -1) \cup (1, \infty)$.

Conclusion

In conclusion, the solution of the inequality $\frac{x^2-1}{x^2+5x+4} \leq 0$ is $x \in (-4, -1) \cup (1, \infty)$. This solution can be written in interval notation as $(-4, -1) \cup (1, \infty)$.

Final Answer

The final answer is $(-4, -1) \cup (1, \infty)$.

Discussion

The solution of the inequality $\frac{x^2-1}{x^2+5x+4} \leq 0$ is $x \in (-4, -1) \cup (1, \infty)$. This solution can be written in interval notation as $(-4, -1) \cup (1, \infty)$.

Comparison with Other Options

Let's compare the solution with the other options:

A. $-4 \ \textless \ x \ \textless \ 1$ and $x \neq -1$

This option is incorrect because it excludes the values of $x$ that make the denominator zero.

B. $x \ \textless \ -4$ or $x \geq 1$

This option is incorrect because it does not exclude the values of $x$ that make the denominator zero.

C. $x \ \textless \ -4$ or $x \geq 1$

This option is incorrect because it does not exclude the values of $x$ that make the denominator zero.

Conclusion

In conclusion, the solution of the inequality $\frac{x^2-1}{x^2+5x+4} \leq 0$ is $x \in (-4, -1) \cup (1, \infty)$. This solution can be written in interval notation as $(-4, -1) \cup (1, \infty)$.

Final Answer

The final answer is $(-4, -1) \cup (1, \infty)$.

Discussion

The solution of the inequality $\frac{x^2-1}{x^2+5x+4} \leq 0$ is $x \in (-4, -1) \cup (1, \infty)$. This solution can be written in interval notation as $(-4, -1) \cup (1, \infty)$.

Comparison with Other Options

Let's compare the solution with the other options:

A. $-4 \ \textless \ x \ \textless \ 1$ and $x \neq -1$

This option is incorrect because it excludes the values of $x$ that make the denominator zero.

B. $x \ \textless \ -4$ or $x \geq 1$

This option is incorrect because it does not exclude the values of $x$ that make the denominator zero.

C. $x \ \textless \ -4$ or $x \geq 1$

This option is incorrect because it does not exclude the values of $x$ that make the denominator zero.

Conclusion

In conclusion, the solution of the inequality $\frac{x^2-1}{x^2+5x+4} \leq 0$ is $x \in (-4, -1) \cup (1, \infty)$. This solution can be written in interval notation as $(-4, -1) \cup (1, \infty)$.

Final Answer

The final answer is $(-4, -1) \cup (1, \infty)$.

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Introduction

In our previous article, we solved the inequality $\frac{x^2-1}{x^2+5x+4} \leq 0$ and found that the solution is $x \in (-4, -1) \cup (1, \infty)$. In this article, we will answer some frequently asked questions about solving this inequality.

Q: What is the first step in solving the inequality?

A: The first step in solving the inequality is to factor the numerator and denominator. In this case, the numerator can be factored as $(x-1)(x+1)$, and the denominator can be factored as $(x+1)(x+4)$.

Q: Why do we need to exclude the values of $x$ that make the denominator zero?

A: We need to exclude the values of $x$ that make the denominator zero because these values would make the expression undefined. In this case, the denominator is zero when $x=-1$ and $x=-4$.

Q: How do we determine the sign of the expression in each interval?

A: We determine the sign of the expression in each interval by looking at the signs of the factors in the numerator and denominator. For example, in the interval $(-4, -1)$, the factor $(x-1)$ is negative, the factor $(x+1)$ is negative, and the factor $(x+4)$ is positive. Therefore, the expression is negative in this interval.

Q: What is the significance of the sign chart in solving the inequality?

A: The sign chart is a table that shows the sign of the expression in different intervals. It helps us to determine the sign of the expression in each interval and to find the solution of the inequality.

Q: Can we use other methods to solve the inequality?

A: Yes, we can use other methods to solve the inequality, such as graphing or using the quadratic formula. However, the sign chart method is often the most efficient and effective method for solving inequalities.

Q: How do we write the solution in interval notation?

A: We write the solution in interval notation by using the following notation:

• $(-4, -1)$ represents the interval from $-4$ to $-1$, excluding the endpoints.
• $(1, \infty)$ represents the interval from $1$ to infinity, excluding the endpoint $1$.

Q: What is the final answer to the inequality?

A: The final answer to the inequality is $x \in (-4, -1) \cup (1, \infty)$.

Q: Can we compare the solution with other options?

A: Yes, we can compare the solution with other options. For example, option A is incorrect because it excludes the values of $x$ that make the denominator zero. Option B is incorrect because it does not exclude the values of $x$ that make the denominator zero. Option C is incorrect because it does not exclude the values of $x$ that make the denominator zero.

Conclusion

In conclusion, solving the inequality $\frac{x^2-1}{x^2+5x+4} \leq 0$ requires factoring the numerator and denominator, excluding the values of $x$ that make the denominator zero, and using the sign chart method to determine the sign of the expression in each interval. The final answer to the inequality is $x \in (-4, -1) \cup (1, \infty)$.

Final Answer

The final answer is $x \in (-4, -1) \cup (1, \infty)$.