Solve Over The Set Of Complex Numbers:$\[ 4x^2 + 3x^{-2} - 13 = 0 \\]The Solution Set Is \[$\square\$\].

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Introduction

In this article, we will be solving a quadratic equation over the set of complex numbers. The given equation is $4x^2 + 3x^{-2} - 13 = 0$. This equation involves a quadratic term and a reciprocal term, which makes it a bit more challenging to solve. We will use various mathematical techniques and formulas to find the solution set of this equation.

Understanding the Equation

The given equation is a quadratic equation in terms of $x$. However, it also involves a reciprocal term, which is $x^{-2}$. To simplify the equation, we can rewrite it as follows:

$$4x^2 + \frac{3}{x^2} - 13 = 0$$

This equation can be further simplified by multiplying both sides by $x^2$, which gives us:

$$4x^4 + 3 - 13x^2 = 0$$

Rearranging the Equation

To make the equation more manageable, we can rearrange it to get:

$$4x^4 - 13x^2 + 3 = 0$$

This equation is a quadratic equation in terms of $x^2$. We can use the quadratic formula to solve for $x^2$.

Using the Quadratic Formula

The quadratic formula is given by:

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

In this case, we have:

$$a = 4, b = -13, c = 3$$

Substituting these values into the quadratic formula, we get:

$$x^2 = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(4)(3)}}{2(4)}$$

Simplifying this expression, we get:

$$x^2 = \frac{13 \pm \sqrt{169 - 48}}{8}$$

$$x^2 = \frac{13 \pm \sqrt{121}}{8}$$

$$x^2 = \frac{13 \pm 11}{8}$$

Finding the Solutions

Now that we have the values of $x^2$, we can find the solutions for $x$. We have two possible values for $x^2$, which are:

$$x^2 = \frac{13 + 11}{8} = 3$$

$$x^2 = \frac{13 - 11}{8} = \frac{1}{4}$$

Taking the square root of both sides, we get:

$$x = \pm \sqrt{3}$$

$$x = \pm \sqrt{\frac{1}{4}}$$

$$x = \pm \frac{1}{2}$$

Conclusion

In this article, we solved a quadratic equation over the set of complex numbers. The given equation was $4x^2 + 3x^{-2} - 13 = 0$. We used various mathematical techniques and formulas to find the solution set of this equation. The solutions to this equation are $x = \pm \sqrt{3}$ and $x = \pm \frac{1}{2}$.

Final Answer

The final answer is $\boxed{\left( -\frac{1}{2}, \ \frac{1}{2}, \ -\sqrt{3}, \ \sqrt{3}\right)}$

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Introduction

In our previous article, we solved a quadratic equation over the set of complex numbers. The given equation was $4x^2 + 3x^{-2} - 13 = 0$. We used various mathematical techniques and formulas to find the solution set of this equation. In this article, we will answer some frequently asked questions related to this topic.

Q&A

Q: What is the difference between a quadratic equation and a quadratic expression?

A: A quadratic equation is an equation that involves a quadratic expression, which is a polynomial of degree two. In other words, a quadratic equation is an equation that can be written in the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants.

Q: How do you solve a quadratic equation?

A: There are several methods to solve a quadratic equation, including factoring, completing the square, and using the quadratic formula. The quadratic formula is given by:

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

Q: What is the quadratic formula?

A: The quadratic formula is a formula that is used to solve quadratic equations. It is given by:

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

Q: How do you use the quadratic formula?

A: To use the quadratic formula, you need to identify the values of $a$, $b$, and $c$ in the quadratic equation. Then, you can plug these values into the quadratic formula and simplify to find the solutions.

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

A: A real solution is a solution that is a real number, while a complex solution is a solution that is a complex number. In other words, a real solution is a solution that can be written in the form $x = a$, where $a$ is a real number, while a complex solution is a solution that can be written in the form $x = a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit.

Q: How do you find the complex solutions of a quadratic equation?

A: To find the complex solutions of a quadratic equation, you can use the quadratic formula and simplify to find the solutions. If the solutions are complex, they will be in the form $x = a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit.

Q: What is the significance of the imaginary unit $i$?

A: The imaginary unit $i$ is a complex number that satisfies the equation $i^2 = -1$. It is used to represent complex numbers and is an essential part of complex analysis.

Q: How do you simplify complex expressions?

A: To simplify complex expressions, you can use the rules of arithmetic, such as the distributive property and the commutative property. You can also use the fact that $i^2 = -1$ to simplify expressions involving $i$.

Conclusion

In this article, we answered some frequently asked questions related to solving quadratic equations over the set of complex numbers. We discussed the difference between a quadratic equation and a quadratic expression, how to solve a quadratic equation, and the significance of the imaginary unit $i$. We also provided some tips on how to simplify complex expressions.

Final Answer

The final answer is $\boxed{\left( -\frac{1}{2}, \ \frac{1}{2}, \ -\sqrt{3}, \ \sqrt{3}\right)}$