Solve The Inequality:$\[ 2x^3 - 5x^2 \leq 25x \\]The Solution Set Is \[$\square\$\]. (Type Your Answer In Interval Notation.)

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Introduction

In this article, we will delve into solving the given inequality $2x^3 - 5x^2 \leq 25x$. This involves manipulating the inequality to isolate the variable $x$ and determine the solution set in interval notation. We will use algebraic techniques to simplify the inequality and identify the critical points that divide the number line into intervals where the inequality is either true or false.

Step 1: Move all terms to one side of the inequality

To begin solving the inequality, we need to move all terms to one side of the inequality. This will allow us to factor the left-hand side and identify the critical points.

${ 2x^3 - 5x^2 - 25x \leq 0 }$

Step 2: Factor out the common term

We can factor out the common term $x$ from the left-hand side of the inequality.

${ x(2x^2 - 5x - 25) \leq 0 }$

Step 3: Factor the quadratic expression

Now, we need to factor the quadratic expression $2x^2 - 5x - 25$. We can use the quadratic formula to find the roots of the quadratic equation $2x^2 - 5x - 25 = 0$.

${ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} }$

Substituting the values $a = 2$, $b = -5$, and $c = -25$, we get:

${ x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(-25)}}{2(2)} }$

${ x = \frac{5 \pm \sqrt{25 + 200}}{4} }$

${ x = \frac{5 \pm \sqrt{225}}{4} }$

${ x = \frac{5 \pm 15}{4} }$

This gives us two possible values for $x$:

${ x = \frac{5 + 15}{4} = \frac{20}{4} = 5 }$

${ x = \frac{5 - 15}{4} = \frac{-10}{4} = -\frac{5}{2} }$

Step 4: Factor the quadratic expression

Now that we have the roots of the quadratic equation, we can factor the quadratic expression $2x^2 - 5x - 25$ as:

${ 2x^2 - 5x - 25 = (x - 5)(2x + 5) }$

Step 5: Rewrite the inequality with the factored quadratic expression

Substituting the factored quadratic expression back into the inequality, we get:

${ x(x - 5)(2x + 5) \leq 0 }$

Step 6: Identify the critical points

The critical points are the values of $x$ that make the expression $x(x - 5)(2x + 5)$ equal to zero. These points are $x = 0$, $x = 5$, and $x = -\frac{5}{2}$.

Step 7: Test the intervals

To determine the solution set, we need to test the intervals defined by the critical points. The intervals are $(-\infty, -\frac{5}{2})$, $(-\frac{5}{2}, 0)$, $(0, 5)$, and $(5, \infty)$.

Step 8: Determine the solution set

After testing the intervals, we find that the solution set is $(-\infty, -\frac{5}{2}) \cup [0, 5]$.

Conclusion

In this article, we solved the inequality $2x^3 - 5x^2 \leq 25x$ by manipulating the inequality to isolate the variable $x$ and determining the solution set in interval notation. We used algebraic techniques to simplify the inequality and identify the critical points that divide the number line into intervals where the inequality is either true or false. The solution set is $(-\infty, -\frac{5}{2}) \cup [0, 5]$.

Final Answer

The final answer is $\boxed{(-\infty, -\frac{5}{2}) \cup [0, 5]}$.

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Introduction

In our previous article, we solved the inequality $2x^3 - 5x^2 \leq 25x$ and determined the solution set in interval notation. In this article, we will answer some frequently asked questions related to solving the inequality.

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

A: The first step in solving the inequality is to move all terms to one side of the inequality. This will allow us to factor the left-hand side and identify the critical points.

Q: How do I factor the quadratic expression?

A: To factor the quadratic expression, we need to find the roots of the quadratic equation. We can use the quadratic formula to find the roots of the quadratic equation.

Q: What are the critical points?

A: The critical points are the values of $x$ that make the expression $x(x - 5)(2x + 5)$ equal to zero. These points are $x = 0$, $x = 5$, and $x = -\frac{5}{2}$.

Q: How do I test the intervals?

A: To test the intervals, we need to choose a test point from each interval and substitute it into the inequality. If the inequality is true for the test point, then the entire interval is part of the solution set.

Q: What is the solution set?

A: The solution set is $(-\infty, -\frac{5}{2}) \cup [0, 5]$.

Q: How do I graph the solution set?

A: To graph the solution set, we need to plot the critical points on the number line and shade the intervals that are part of the solution set.

Q: What are some common mistakes to avoid when solving inequalities?

A: Some common mistakes to avoid when solving inequalities include:

• Not moving all terms to one side of the inequality
• Not factoring the quadratic expression correctly
• Not identifying the critical points correctly
• Not testing the intervals correctly

Q: How do I check my work?

A: To check your work, you can substitute a test point from each interval into the inequality and verify that the inequality is true.

Q: What are some real-world applications of solving inequalities?

A: Solving inequalities has many real-world applications, including:

• Modeling population growth
• Modeling financial investments
• Modeling physical systems

Conclusion

In this article, we answered some frequently asked questions related to solving the inequality $2x^3 - 5x^2 \leq 25x$. We covered topics such as factoring the quadratic expression, identifying the critical points, testing the intervals, and graphing the solution set. We also discussed some common mistakes to avoid and how to check your work.

Final Answer

The final answer is $\boxed{(-\infty, -\frac{5}{2}) \cup [0, 5]}$.

Additional Resources

• For more information on solving inequalities, see our previous article on solving the inequality $2x^3 - 5x^2 \leq 25x$.
• For more information on graphing the solution set, see our article on graphing the solution set of an inequality.
• For more information on real-world applications of solving inequalities, see our article on real-world applications of solving inequalities.