Solve For $y$.$4y^2 = -20y - 25$If There Is More Than One Solution, Separate Them With Commas. If There Is No Solution, Click On No Solution.
Introduction
Solving quadratic equations is a fundamental concept in mathematics, and it is essential to understand how to approach these types of problems. In this article, we will focus on solving the quadratic equation 4y^2 = -20y - 25. We will use various methods to find the solutions, and we will also discuss the significance of these solutions in real-world applications.
Understanding the Equation
The given equation is a quadratic equation in the form of ay^2 + by + c = 0, where a = 4, b = -20, and c = -25. To solve this equation, we can use various methods such as factoring, completing the square, or using the quadratic formula.
Factoring the Equation
One way to solve the equation is by factoring it. However, in this case, the equation does not factor easily, and we need to use other methods to find the solutions.
Completing the Square
Another method to solve the equation is by completing the square. This method involves manipulating the equation to create a perfect square trinomial. We can start by dividing both sides of the equation by 4 to simplify it:
y^2 + 5y + 6.25 = 0
Next, we can add 6.25 to both sides of the equation to create a perfect square trinomial:
y^2 + 5y = -6.25
Now, we can add (5/2)^2 = 6.25 to both sides of the equation to complete the square:
y^2 + 5y + 6.25 = 0 + 6.25
This simplifies to:
(y + 2.5)^2 = 0
Solving for y
Now that we have completed the square, we can solve for y by taking the square root of both sides of the equation:
y + 2.5 = 0
Subtracting 2.5 from both sides of the equation gives us:
y = -2.5
Checking the Solution
To check our solution, we can plug it back into the original equation:
4y^2 = -20y - 25
Substituting y = -2.5 into the equation gives us:
4(-2.5)^2 = -20(-2.5) - 25
Expanding and simplifying the equation gives us:
4(6.25) = 50 - 25
25 = 25
This confirms that our solution is correct.
Conclusion
In this article, we solved the quadratic equation 4y^2 = -20y - 25 using the method of completing the square. We found that the solution to the equation is y = -2.5. We also checked our solution by plugging it back into the original equation to confirm that it is correct.
Real-World Applications
Quadratic equations have numerous real-world applications, including physics, engineering, and economics. For example, the motion of an object under the influence of gravity can be modeled using quadratic equations. In finance, quadratic equations can be used to model the growth of investments over time.
Final Answer
The final answer to the equation 4y^2 = -20y - 25 is y = -2.5.
Introduction
In our previous article, we solved the quadratic equation 4y^2 = -20y - 25 using the method of completing the square. We found that the solution to the equation is y = -2.5. In this article, we will answer some frequently asked questions related to the solution of the equation.
Q&A
Q: What is the significance of the solution y = -2.5?
A: The solution y = -2.5 represents the value of y that satisfies the equation 4y^2 = -20y - 25. In other words, it is the value of y that makes the equation true.
Q: How do I check if the solution is correct?
A: To check if the solution is correct, you can plug it back into the original equation. If the equation holds true, then the solution is correct.
Q: Can I use other methods to solve the equation?
A: Yes, you can use other methods to solve the equation, such as factoring or using the quadratic formula. However, the method of completing the square is a useful technique to have in your toolkit.
Q: What if the equation has no solution?
A: If the equation has no solution, it means that there is no value of y that satisfies the equation. In this case, the equation is said to be inconsistent.
Q: Can I use quadratic equations to model real-world problems?
A: Yes, quadratic equations can be used to model a wide range of real-world problems, including physics, engineering, and economics.
Q: How do I apply the solution to a real-world problem?
A: To apply the solution to a real-world problem, you need to understand the context of the problem and how the solution relates to it. For example, if you are modeling the motion of an object under the influence of gravity, the solution y = -2.5 might represent the height of the object at a particular time.
Q: Can I use technology to solve quadratic equations?
A: Yes, you can use technology such as calculators or computer software to solve quadratic equations. However, it is still important to understand the underlying mathematics and how to apply the solution to real-world problems.
Conclusion
In this article, we answered some frequently asked questions related to the solution of the quadratic equation 4y^2 = -20y - 25. We hope that this article has been helpful in clarifying any doubts you may have had about the solution.
Real-World Applications
Quadratic equations have numerous real-world applications, including physics, engineering, and economics. For example, the motion of an object under the influence of gravity can be modeled using quadratic equations. In finance, quadratic equations can be used to model the growth of investments over time.
Final Answer
The final answer to the equation 4y^2 = -20y - 25 is y = -2.5.
Additional Resources
If you want to learn more about quadratic equations and how to apply them to real-world problems, we recommend checking out the following resources:
- Khan Academy: Quadratic Equations
- Mathway: Quadratic Equations
- Wolfram Alpha: Quadratic Equations
We hope that this article has been helpful in clarifying any doubts you may have had about the solution of the quadratic equation 4y^2 = -20y - 25.