Solve The Inequality:$ 3x^2 - 4x + 2 \leq X^2 + X + 6 $

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Introduction

In this article, we will delve into solving the given inequality, $3x^2 - 4x + 2 \leq x^2 + x + 6$. This involves manipulating the inequality to isolate the variable, $x$, and determining the values of $x$ that satisfy the given condition. We will use algebraic techniques to simplify the inequality and find the solution set.

Understanding the Inequality

The given inequality is a quadratic inequality, which involves a quadratic expression on both sides of the inequality sign. The left-hand side of the inequality is $3x^2 - 4x + 2$, while the right-hand side is $x^2 + x + 6$. Our goal is to find the values of $x$ that make the inequality true.

Step 1: Simplify the Inequality

To simplify the inequality, we can start by combining like terms on both sides of the inequality sign. This involves subtracting $x^2 + x + 6$ from both sides of the inequality.

3x^2 - 4x + 2 \leq x^2 + x + 6

Subtracting $x^2 + x + 6$ from both sides gives us:

3x^2 - 4x + 2 - (x^2 + x + 6) \leq 0

Simplifying the left-hand side of the inequality, we get:

2x^2 - 5x - 4 \leq 0

Step 2: Factor the Quadratic Expression

The next step is to factor the quadratic expression on the left-hand side of the inequality. We can factor the expression as follows:

2x^2 - 5x - 4 = (2x + 1)(x - 4)

Therefore, the inequality can be written as:

(2x + 1)(x - 4) \leq 0

Step 3: Find the Critical Points

To find the critical points of the inequality, we need to set each factor equal to zero and solve for $x$. Setting $2x + 1 = 0$, we get:

2x + 1 = 0

Solving for $x$, we get:

x = -\frac{1}{2}

Setting $x - 4 = 0$, we get:

x - 4 = 0

Solving for $x$, we get:

x = 4

Step 4: Determine the Solution Set

To determine the solution set of the inequality, we need to test the intervals defined by the critical points. The critical points are $x = -\frac{1}{2}$ and $x = 4$. We can test the intervals $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 4)$, and $(4, \infty)$.

Interval $(-\infty, -\frac{1}{2})$

Let's test the interval $(-\infty, -\frac{1}{2})$ by choosing a value of $x$ in this interval, say $x = -1$. Plugging this value into the inequality, we get:

(2(-1) + 1)(-1 - 4) \leq 0

Simplifying the expression, we get:

(-1)(-5) \leq 0

Since $5 \leq 0$ is not true, the inequality is not satisfied in this interval.

Interval $(-\frac{1}{2}, 4)$

Let's test the interval $(-\frac{1}{2}, 4)$ by choosing a value of $x$ in this interval, say $x = 0$. Plugging this value into the inequality, we get:

(2(0) + 1)(0 - 4) \leq 0

Simplifying the expression, we get:

(1)(-4) \leq 0

Since $-4 \leq 0$ is true, the inequality is satisfied in this interval.

Interval $(4, \infty)$

Let's test the interval $(4, \infty)$ by choosing a value of $x$ in this interval, say $x = 5$. Plugging this value into the inequality, we get:

(2(5) + 1)(5 - 4) \leq 0

Simplifying the expression, we get:

(11)(1) \leq 0

Since $11 \leq 0$ is not true, the inequality is not satisfied in this interval.

Conclusion

In conclusion, the solution set of the inequality $3x^2 - 4x + 2 \leq x^2 + x + 6$ is the interval $(-\infty, -\frac{1}{2}] \cup [4, \infty)$. This means that the inequality is satisfied for all values of $x$ less than or equal to $-\frac{1}{2}$ and for all values of $x$ greater than or equal to $4$.

Final Answer

The final answer is $\boxed{(-\infty, -\frac{1}{2}] \cup [4, \infty)}$.

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Introduction

In this article, we will delve into solving the given inequality, $3x^2 - 4x + 2 \leq x^2 + x + 6$. This involves manipulating the inequality to isolate the variable, $x$, and determining the values of $x$ that satisfy the given condition. We will use algebraic techniques to simplify the inequality and find the solution set.

Understanding the Inequality

The given inequality is a quadratic inequality, which involves a quadratic expression on both sides of the inequality sign. The left-hand side of the inequality is $3x^2 - 4x + 2$, while the right-hand side is $x^2 + x + 6$. Our goal is to find the values of $x$ that make the inequality true.

Step 1: Simplify the Inequality

To simplify the inequality, we can start by combining like terms on both sides of the inequality sign. This involves subtracting $x^2 + x + 6$ from both sides of the inequality.

3x^2 - 4x + 2 \leq x^2 + x + 6

Subtracting $x^2 + x + 6$ from both sides gives us:

3x^2 - 4x + 2 - (x^2 + x + 6) \leq 0

Simplifying the left-hand side of the inequality, we get:

2x^2 - 5x - 4 \leq 0

Step 2: Factor the Quadratic Expression

The next step is to factor the quadratic expression on the left-hand side of the inequality. We can factor the expression as follows:

2x^2 - 5x - 4 = (2x + 1)(x - 4)

Therefore, the inequality can be written as:

(2x + 1)(x - 4) \leq 0

Step 3: Find the Critical Points

To find the critical points of the inequality, we need to set each factor equal to zero and solve for $x$. Setting $2x + 1 = 0$, we get:

2x + 1 = 0

Solving for $x$, we get:

x = -\frac{1}{2}

Setting $x - 4 = 0$, we get:

x - 4 = 0

Solving for $x$, we get:

x = 4

Step 4: Determine the Solution Set

To determine the solution set of the inequality, we need to test the intervals defined by the critical points. The critical points are $x = -\frac{1}{2}$ and $x = 4$. We can test the intervals $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 4)$, and $(4, \infty)$.

Interval $(-\infty, -\frac{1}{2})$

Let's test the interval $(-\infty, -\frac{1}{2})$ by choosing a value of $x$ in this interval, say $x = -1$. Plugging this value into the inequality, we get:

(2(-1) + 1)(-1 - 4) \leq 0

Simplifying the expression, we get:

(-1)(-5) \leq 0

Since $5 \leq 0$ is not true, the inequality is not satisfied in this interval.

Interval $(-\frac{1}{2}, 4)$

Let's test the interval $(-\frac{1}{2}, 4)$ by choosing a value of $x$ in this interval, say $x = 0$. Plugging this value into the inequality, we get:

(2(0) + 1)(0 - 4) \leq 0

Simplifying the expression, we get:

(1)(-4) \leq 0

Since $-4 \leq 0$ is true, the inequality is satisfied in this interval.

Interval $(4, \infty)$

Let's test the interval $(4, \infty)$ by choosing a value of $x$ in this interval, say $x = 5$. Plugging this value into the inequality, we get:

(2(5) + 1)(5 - 4) \leq 0

Simplifying the expression, we get:

(11)(1) \leq 0

Since $11 \leq 0$ is not true, the inequality is not satisfied in this interval.

Conclusion

In conclusion, the solution set of the inequality $3x^2 - 4x + 2 \leq x^2 + x + 6$ is the interval $(-\infty, -\frac{1}{2}] \cup [4, \infty)$. This means that the inequality is satisfied for all values of $x$ less than or equal to $-\frac{1}{2}$ and for all values of $x$ greater than or equal to $4$.

Q&A

Q: What is the solution set of the inequality $3x^2 - 4x + 2 \leq x^2 + x + 6$?

A: The solution set of the inequality is the interval $(-\infty, -\frac{1}{2}] \cup [4, \infty)$.

Q: How do I find the critical points of the inequality?

A: To find the critical points of the inequality, you need to set each factor equal to zero and solve for $x$. In this case, the critical points are $x = -\frac{1}{2}$ and $x = 4$.

Q: How do I determine the solution set of the inequality?

A: To determine the solution set of the inequality, you need to test the intervals defined by the critical points. In this case, we tested the intervals $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 4)$, and $(4, \infty)$.

Q: What is the significance of the critical points in the solution set?

A: The critical points are the values of $x$ that make the inequality true or false. In this case, the critical points are $x = -\frac{1}{2}$ and $x = 4$, which divide the solution set into three intervals.

Q: How do I choose a value of $x$ to test in each interval?

A: To choose a value of $x$ to test in each interval, you can pick a value that is easy to work with and that is in the interval. For example, in the interval $(-\infty, -\frac{1}{2})$, you can choose $x = -1$.

Q: What is the final answer to the inequality?

A: The final answer to the inequality is $\boxed{(-\infty, -\frac{1}{2}] \cup [4, \infty)}$.

Final Answer

The final answer is $\boxed{(-\infty, -\frac{1}{2}] \cup [4, \infty)}$.