Which Equation Is Equivalent To $\[ F(x) = 16x^4 - 81 = 0 \\]?A. \[$(4x^2 + 9)(4x^2 - 9) = 0\$\]Select All Of The Zeros Of The Function:B. \[$\pm \frac{3}{2}\$\]C. \[$\pm \frac{3}{2}i\$\]D. \[$\pm

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Introduction

In this article, we will explore the process of solving a quartic equation, specifically the equation f(x)=16x4−81=0f(x) = 16x^4 - 81 = 0. We will break down the solution into manageable steps and provide a clear explanation of each step. By the end of this article, you will have a thorough understanding of how to solve this type of equation and be able to apply the same techniques to other similar equations.

Step 1: Factor the Equation

The first step in solving the equation f(x)=16x4−81=0f(x) = 16x^4 - 81 = 0 is to factor the left-hand side of the equation. We can start by recognizing that 16x416x^4 is a perfect square, which can be written as (4x2)2(4x^2)^2. Similarly, 8181 is also a perfect square, which can be written as 929^2. Therefore, we can rewrite the equation as:

f(x)=(4x2)2−92=0f(x) = (4x^2)^2 - 9^2 = 0

Now, we can use the difference of squares formula to factor the left-hand side of the equation:

(4x2−9)(4x2+9)=0(4x^2 - 9)(4x^2 + 9) = 0

This gives us two separate equations:

4x2−9=0or4x2+9=04x^2 - 9 = 0 \quad \text{or} \quad 4x^2 + 9 = 0

Step 2: Solve the First Equation

The first equation is a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0. In this case, we have:

4x2−9=04x^2 - 9 = 0

We can solve this equation by adding 99 to both sides and then dividing by 44:

4x2=94x^2 = 9

x2=94x^2 = \frac{9}{4}

x=±94x = \pm \sqrt{\frac{9}{4}}

x=±32x = \pm \frac{3}{2}

Therefore, the solutions to the first equation are x=32x = \frac{3}{2} and x=−32x = -\frac{3}{2}.

Step 3: Solve the Second Equation

The second equation is also a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c = 0. In this case, we have:

4x2+9=04x^2 + 9 = 0

We can solve this equation by subtracting 99 from both sides and then dividing by 44:

4x2=−94x^2 = -9

x2=−94x^2 = -\frac{9}{4}

x=±−94x = \pm \sqrt{-\frac{9}{4}}

x=±32ix = \pm \frac{3}{2}i

Therefore, the solutions to the second equation are x=32ix = \frac{3}{2}i and x=−32ix = -\frac{3}{2}i.

Conclusion

In this article, we have solved the quartic equation f(x)=16x4−81=0f(x) = 16x^4 - 81 = 0 by factoring the left-hand side of the equation and then solving the resulting quadratic equations. We have found that the solutions to the equation are x=32x = \frac{3}{2}, x=−32x = -\frac{3}{2}, x=32ix = \frac{3}{2}i, and x=−32ix = -\frac{3}{2}i.

Answer Key

Based on the solutions we have found, we can select the correct answer from the options provided:

  • A. (4x2+9)(4x2−9)=0(4x^2 + 9)(4x^2 - 9) = 0 is the correct factorization of the equation.
  • B. ±32\pm \frac{3}{2} are the real solutions to the equation.
  • C. ±32i\pm \frac{3}{2}i are the complex solutions to the equation.

Therefore, the correct answer is:

  • A. (4x2+9)(4x2−9)=0(4x^2 + 9)(4x^2 - 9) = 0
  • B. ±32\pm \frac{3}{2}
  • C. ±32i\pm \frac{3}{2}i

Discussion

The equation f(x)=16x4−81=0f(x) = 16x^4 - 81 = 0 is a quartic equation, which means it has four solutions. In this article, we have solved the equation by factoring the left-hand side and then solving the resulting quadratic equations. We have found that the solutions to the equation are x=32x = \frac{3}{2}, x=−32x = -\frac{3}{2}, x=32ix = \frac{3}{2}i, and x=−32ix = -\frac{3}{2}i.

The solutions to the equation can be classified into two categories: real solutions and complex solutions. The real solutions are x=32x = \frac{3}{2} and x=−32x = -\frac{3}{2}, while the complex solutions are x=32ix = \frac{3}{2}i and x=−32ix = -\frac{3}{2}i.

Q: What is a quartic equation?

A: A quartic equation is a polynomial equation of degree four, which means it has four solutions. It is typically written in the form ax4+bx3+cx2+dx+e=0ax^4 + bx^3 + cx^2 + dx + e = 0, where aa, bb, cc, dd, and ee are constants.

Q: How do I solve a quartic equation?

A: Solving a quartic equation can be a complex process, but it typically involves factoring the left-hand side of the equation and then solving the resulting quadratic equations. In some cases, it may be possible to use numerical methods or approximation techniques to find the solutions.

Q: What are the different types of solutions to a quartic equation?

A: The solutions to a quartic equation can be classified into two categories: real solutions and complex solutions. Real solutions are solutions that can be expressed as a real number, while complex solutions are solutions that involve imaginary numbers.

Q: How do I determine if a solution is real or complex?

A: To determine if a solution is real or complex, you can use the discriminant of the quadratic equation that results from factoring the quartic equation. If the discriminant is positive, the solution is real. If the discriminant is negative, the solution is complex.

Q: What is the difference between a real solution and a complex solution?

A: A real solution is a solution that can be expressed as a real number, while a complex solution is a solution that involves imaginary numbers. For example, the solution x=2x = 2 is a real solution, while the solution x=2ix = 2i is a complex solution.

Q: Can I use numerical methods to solve a quartic equation?

A: Yes, it is possible to use numerical methods to solve a quartic equation. Numerical methods involve using an iterative process to approximate the solutions to the equation. This can be useful when the equation is too complex to solve analytically.

Q: What are some common numerical methods for solving quartic equations?

A: Some common numerical methods for solving quartic equations include the Newton-Raphson method, the bisection method, and the secant method. These methods involve using an iterative process to approximate the solutions to the equation.

Q: How do I choose the best numerical method for solving a quartic equation?

A: The choice of numerical method depends on the specific equation and the desired level of accuracy. Some methods are more efficient than others, while others may be more accurate. It is often helpful to try out different methods and compare the results to determine which one is best.

Q: Can I use a computer algebra system (CAS) to solve a quartic equation?

A: Yes, many computer algebra systems (CAS) are capable of solving quartic equations. CAS software can often solve equations analytically, but may also be able to use numerical methods to approximate the solutions.

Q: What are some common applications of quartic equations?

A: Quartic equations have many applications in mathematics, science, and engineering. Some common applications include:

  • Physics: Quartic equations are used to model the motion of objects under the influence of gravity and other forces.
  • Engineering: Quartic equations are used to design and optimize systems, such as bridges and buildings.
  • Computer Science: Quartic equations are used in computer graphics and game development to create realistic simulations.
  • Mathematics: Quartic equations are used to study the properties of polynomials and to develop new mathematical techniques.

Conclusion

Quartic equations are a fundamental concept in mathematics, with many applications in science, engineering, and computer science. By understanding how to solve quartic equations, you can gain a deeper appreciation for the underlying mathematics and develop new skills and techniques. Whether you are a student, a researcher, or a practitioner, quartic equations are an essential tool for solving complex problems and making new discoveries.