Complex Numbers and Quadratic Equations

Oct 13, 2025 by windailys.com 74 views

Introduction

Quadratic equations are a fundamental concept in mathematics, and they can be solved using various methods, including factoring, the quadratic formula, and completing the square. However, when the solutions to a quadratic equation involve complex numbers, it can be challenging to determine the other solution. In this article, we will explore how to find the other solution to a quadratic equation when one solution is given as a complex number.

Complex Numbers and Quadratic Equations

Complex numbers are numbers that can be expressed in the form $a + bi$ , where $a$ and $b$ are real numbers and $i$ is the imaginary unit, which satisfies the equation $i^2 = -1$ . Quadratic equations with complex solutions can be written in the form $ax^2 + bx + c = 0$ , where $a$ , $b$ , and $c$ are real numbers.

The Conjugate Root Theorem

The conjugate root theorem states that if a polynomial equation with real coefficients has a complex root, then its conjugate is also a root of the equation. In other words, if $z = a + bi$ is a root of the equation, then its conjugate $\overline{z} = a - bi$ is also a root.

Applying the Conjugate Root Theorem

In the given problem, we are told that one solution to the quadratic equation is $x = -2i$ . Since the equation has real coefficients, we can apply the conjugate root theorem to find the other solution. The conjugate of $-2i$ is $2i$ , which is option A.

Why the Other Options are Incorrect

Let's examine why the other options are incorrect.

Option B: $x = -3i$ is not the conjugate of $-2i$ , so it cannot be the other solution.
Option C: $x = -3$ is a real number, not a complex number, so it cannot be the other solution.
Option D: $x = 2$ is also a real number, not a complex number, so it cannot be the other solution.

Conclusion

In conclusion, when one solution to a quadratic equation is given as a complex number, we can use the conjugate root theorem to find the other solution. In this case, the other solution is $x = 2i$ , which is option A.

Example Problems

Here are a few example problems to illustrate the concept:

If one solution to the quadratic equation $x^2 + 4x + 5 = 0$ is $x = -2i$ , what is the other solution?
If one solution to the quadratic equation $x^2 - 6x + 10 = 0$ is $x = 3i$ , what is the other solution?

Solutions to Example Problems

The other solution to the quadratic equation $x^2 + 4x + 5 = 0$ is $x = 2i$ .
The other solution to the quadratic equation $x^2 - 6x + 10 = 0$ is $x = -3i$ .

Final Thoughts

Introduction

In our previous article, we explored how to find the other solution to a quadratic equation when one solution is given as a complex number. We used the conjugate root theorem to find the other solution and provided example problems to illustrate the concept. In this article, we will answer some frequently asked questions about quadratic equations with complex solutions.

Q: What is the conjugate root theorem?

A: The conjugate root theorem states that if a polynomial equation with real coefficients has a complex root, then its conjugate is also a root of the equation. In other words, if $z = a + bi$ is a root of the equation, then its conjugate $\overline{z} = a - bi$ is also a root.

Q: How do I apply the conjugate root theorem to find the other solution?

A: To apply the conjugate root theorem, you need to find the conjugate of the given complex solution. The conjugate of a complex number $z = a + bi$ is $\overline{z} = a - bi$ . Once you have found the conjugate, you can use it as the other solution.

Q: What if the given solution is not in the form $a + bi$ ?

A: If the given solution is not in the form $a + bi$ , you need to rewrite it in this form. For example, if the given solution is $-2i$ , you can rewrite it as $0 - 2i$ , which is in the form $a + bi$ .

Q: Can I use the conjugate root theorem to find the other solution if the equation has complex coefficients?

A: No, you cannot use the conjugate root theorem to find the other solution if the equation has complex coefficients. The conjugate root theorem only applies to polynomial equations with real coefficients.

Q: What if the equation has no real solutions?

A: If the equation has no real solutions, it means that the equation has only complex solutions. In this case, you can use the conjugate root theorem to find the other solution.

Q: Can I use the conjugate root theorem to find the other solution if the equation has multiple complex solutions?

A: Yes, you can use the conjugate root theorem to find the other solution if the equation has multiple complex solutions. However, you need to find the conjugate of each complex solution separately.

Q: How do I know if the equation has complex solutions?

A: You can use the discriminant to determine if the equation has complex solutions. The discriminant is given by the formula $b^2 - 4ac$ . If the discriminant is negative, the equation has complex solutions.

Q: What if the equation has complex solutions, but the given solution is not in the form $a + bi$ ?

A: If the equation has complex solutions, but the given solution is not in the form $a + bi$ , you need to rewrite it in this form. For example, if the given solution is $-2i$ , you can rewrite it as $0 - 2i$ , which is in the form $a + bi$ .

Conclusion

In this article, we have answered some frequently asked questions about quadratic equations with complex solutions. We have used the conjugate root theorem to find the other solution and provided examples to illustrate the concept. We hope that this article has been helpful in understanding the concept of complex solutions to quadratic equations.

Example Problems

Here are a few example problems to illustrate the concept:

If one solution to the quadratic equation $x^2 + 4x + 5 = 0$ is $x = -2i$ , what is the other solution?
If one solution to the quadratic equation $x^2 - 6x + 10 = 0$ is $x = 3i$ , what is the other solution?
If one solution to the quadratic equation $x^2 + 2x + 2 = 0$ is $x = -i$ , what is the other solution?

Solutions to Example Problems

The other solution to the quadratic equation $x^2 + 4x + 5 = 0$ is $x = 2i$ .
The other solution to the quadratic equation $x^2 - 6x + 10 = 0$ is $x = -3i$ .
The other solution to the quadratic equation $x^2 + 2x + 2 = 0$ is $x = i$ .

Final Thoughts

In this article, we have explored some frequently asked questions about quadratic equations with complex solutions. We have used the conjugate root theorem to find the other solution and provided examples to illustrate the concept. We hope that this article has been helpful in understanding the concept of complex solutions to quadratic equations.