Which System Of Equations Can Be Used To Find The Roots Of The Equation 4 X 5 − 12 X 4 + 6 X = 5 X 3 − 2 X 4x^5 - 12x^4 + 6x = 5x^3 - 2x 4 X 5 − 12 X 4 + 6 X = 5 X 3 − 2 X ?A. { Y = − 4 X 5 + 12 X 4 − 6 X Y = 5 X 3 − 2 X \begin{cases} y = -4x^5 + 12x^4 - 6x \\ y = 5x^3 - 2x \end{cases} { Y = − 4 X 5 + 12 X 4 − 6 X Y = 5 X 3 − 2 X ​ B. $\begin{cases} y = 4x^5 - 12x^4 - 5x^3 + 8x \ y

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Introduction

Solving systems of equations is a fundamental concept in mathematics, particularly in algebra and calculus. It involves finding the values of variables that satisfy multiple equations simultaneously. In this article, we will explore which system of equations can be used to find the roots of the equation 4x5−12x4+6x=5x3−2x4x^5 - 12x^4 + 6x = 5x^3 - 2x. We will examine two possible systems of equations and determine which one is correct.

Understanding the Problem

The given equation is a fifth-degree polynomial equation, which can be challenging to solve directly. However, we can use a system of equations to find the roots of the equation. The idea is to rewrite the equation as a system of two equations, where one equation is the original equation and the other equation is a modified version of the original equation.

System A:

The first system of equations is:

{y=−4x5+12x4−6xy=5x3−2x\begin{cases} y = -4x^5 + 12x^4 - 6x \\ y = 5x^3 - 2x \end{cases}

To determine if this system is correct, we need to check if the two equations are equivalent. We can do this by setting the two equations equal to each other and simplifying.

−4x5+12x4−6x=5x3−2x-4x^5 + 12x^4 - 6x = 5x^3 - 2x

Expanding the left-hand side, we get:

−4x5+12x4−6x=−4x5+12x4−6x-4x^5 + 12x^4 - 6x = -4x^5 + 12x^4 - 6x

This simplifies to:

0=00 = 0

This means that the two equations are equivalent, and the system is correct.

System B:

The second system of equations is:

{y=4x5−12x4−5x3+8xy=5x3−2x\begin{cases} y = 4x^5 - 12x^4 - 5x^3 + 8x \\ y = 5x^3 - 2x \end{cases}

To determine if this system is correct, we need to check if the two equations are equivalent. We can do this by setting the two equations equal to each other and simplifying.

4x5−12x4−5x3+8x=5x3−2x4x^5 - 12x^4 - 5x^3 + 8x = 5x^3 - 2x

Expanding the left-hand side, we get:

4x5−12x4−5x3+8x=5x3−2x4x^5 - 12x^4 - 5x^3 + 8x = 5x^3 - 2x

This simplifies to:

4x5−12x4−10x3+8x=04x^5 - 12x^4 - 10x^3 + 8x = 0

This means that the two equations are not equivalent, and the system is not correct.

Conclusion

In conclusion, the correct system of equations to find the roots of the equation 4x5−12x4+6x=5x3−2x4x^5 - 12x^4 + 6x = 5x^3 - 2x is:

{y=−4x5+12x4−6xy=5x3−2x\begin{cases} y = -4x^5 + 12x^4 - 6x \\ y = 5x^3 - 2x \end{cases}

This system is correct because the two equations are equivalent, and we can use it to find the roots of the equation.

Why is System A Correct?

System A is correct because the two equations are equivalent. When we set the two equations equal to each other, we get:

−4x5+12x4−6x=5x3−2x-4x^5 + 12x^4 - 6x = 5x^3 - 2x

This simplifies to:

0=00 = 0

This means that the two equations are equivalent, and we can use System A to find the roots of the equation.

Why is System B Incorrect?

System B is incorrect because the two equations are not equivalent. When we set the two equations equal to each other, we get:

4x5−12x4−5x3+8x=5x3−2x4x^5 - 12x^4 - 5x^3 + 8x = 5x^3 - 2x

This simplifies to:

4x5−12x4−10x3+8x=04x^5 - 12x^4 - 10x^3 + 8x = 0

This means that the two equations are not equivalent, and we cannot use System B to find the roots of the equation.

Real-World Applications

Solving systems of equations has many real-world applications, including:

  • Physics and Engineering: Systems of equations are used to model real-world problems, such as the motion of objects and the behavior of electrical circuits.
  • Computer Science: Systems of equations are used in computer graphics, game development, and machine learning.
  • Economics: Systems of equations are used to model economic systems, including supply and demand curves and economic growth models.

Final Thoughts

In conclusion, solving systems of equations is a fundamental concept in mathematics, and it has many real-world applications. By understanding how to solve systems of equations, we can model real-world problems and make predictions about the behavior of complex systems. In this article, we explored which system of equations can be used to find the roots of the equation 4x5−12x4+6x=5x3−2x4x^5 - 12x^4 + 6x = 5x^3 - 2x. We determined that the correct system is:

{y=−4x5+12x4−6xy=5x3−2x\begin{cases} y = -4x^5 + 12x^4 - 6x \\ y = 5x^3 - 2x \end{cases}

This system is correct because the two equations are equivalent, and we can use it to find the roots of the equation.

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Q: What is a system of equations?

A: A system of equations is a set of two or more equations that are related to each other. Each equation in the system is a statement that two or more variables are equal to a certain value.

Q: How do I solve a system of equations?

A: To solve a system of equations, you need to find the values of the variables that satisfy all the equations in the system. There are several methods to solve a system of equations, including substitution, elimination, and graphing.

Q: What is the substitution method?

A: The substitution method is a method of solving a system of equations by substituting the expression for one variable from one equation into the other equation.

Q: What is the elimination method?

A: The elimination method is a method of solving a system of equations by adding or subtracting the equations to eliminate one of the variables.

Q: What is the graphing method?

A: The graphing method is a method of solving a system of equations by graphing the equations on a coordinate plane and finding the point of intersection.

Q: How do I know which method to use?

A: The choice of method depends on the type of system and the variables involved. If the system has two variables and two equations, the substitution or elimination method may be the best choice. If the system has more than two variables or equations, the graphing method may be more suitable.

Q: Can I use a calculator to solve a system of equations?

A: Yes, you can use a calculator to solve a system of equations. Many calculators have built-in functions for solving systems of equations, including the substitution and elimination methods.

Q: How do I check my solution?

A: To check your solution, you need to substitute the values of the variables back into the original equations and verify that they are true.

Q: What if I have a system of equations with no solution?

A: If you have a system of equations with no solution, it means that the equations are inconsistent and there is no value of the variables that satisfies all the equations.

Q: What if I have a system of equations with infinitely many solutions?

A: If you have a system of equations with infinitely many solutions, it means that the equations are dependent and there are many values of the variables that satisfy all the equations.

Q: Can I use a system of equations to model real-world problems?

A: Yes, you can use a system of equations to model real-world problems. Systems of equations are used in many fields, including physics, engineering, computer science, and economics.

Q: How do I apply systems of equations to real-world problems?

A: To apply systems of equations to real-world problems, you need to identify the variables and equations involved, and then use the methods of substitution, elimination, or graphing to solve the system.

Q: What are some common applications of systems of equations?

A: Some common applications of systems of equations include:

  • Physics and Engineering: Systems of equations are used to model the motion of objects and the behavior of electrical circuits.
  • Computer Science: Systems of equations are used in computer graphics, game development, and machine learning.
  • Economics: Systems of equations are used to model economic systems, including supply and demand curves and economic growth models.

Q: How do I choose the right method for solving a system of equations?

A: To choose the right method for solving a system of equations, you need to consider the type of system and the variables involved. If the system has two variables and two equations, the substitution or elimination method may be the best choice. If the system has more than two variables or equations, the graphing method may be more suitable.

Q: Can I use systems of equations to solve optimization problems?

A: Yes, you can use systems of equations to solve optimization problems. Systems of equations can be used to model optimization problems, such as finding the maximum or minimum value of a function.

Q: How do I use systems of equations to solve optimization problems?

A: To use systems of equations to solve optimization problems, you need to identify the variables and equations involved, and then use the methods of substitution, elimination, or graphing to solve the system.

Q: What are some common optimization problems that can be solved using systems of equations?

A: Some common optimization problems that can be solved using systems of equations include:

  • Linear Programming: Systems of equations can be used to solve linear programming problems, which involve finding the maximum or minimum value of a linear function subject to certain constraints.
  • Nonlinear Programming: Systems of equations can be used to solve nonlinear programming problems, which involve finding the maximum or minimum value of a nonlinear function subject to certain constraints.
  • Optimization of Functions: Systems of equations can be used to optimize functions, which involve finding the maximum or minimum value of a function subject to certain constraints.