Question
how many complex
real and non real soloution do the following polynomials equations have
x^2+5=0
x^3+7=0
real and non real soloution do the following polynomials equations have
x^2+5=0
x^3+7=0
Answers
GPT-4o mini
To determine the number of real and non-real (complex) solutions for the given polynomial equations, we can solve each equation step by step.
### 1. \( x^2 + 5 = 0 \)
Rearranging the equation gives:
\[
x^2 = -5
\]
To find the solutions, we take the square root of both sides:
\[
x = \pm \sqrt{-5} = \pm i\sqrt{5}
\]
Since the solutions involve the imaginary unit \(i\), both solutions are complex. Therefore, we can conclude:
- **Number of real solutions:** 0
- **Number of complex (non-real) solutions:** 2 (specifically \(i\sqrt{5}\) and \(-i\sqrt{5}\))
### 2. \( x^3 + 7 = 0 \)
Rearranging the equation gives:
\[
x^3 = -7
\]
Taking the cube root of both sides gives:
\[
x = \sqrt[3]{-7}
\]
The cube root of a negative number is a real number, and there is exactly one real solution. However, there may be complex solutions as well. The cubic polynomial can be factored:
The three solutions to \(x^3 + 7 = 0\) can be found using polar coordinates or other techniques for cubics. In the complex plane, we know that:
1. One real solution: \(x = -\sqrt[3]{7}\)
2. Two non-real solutions can be found by using the polar form. The other solutions will be in the form:
\[
x = -\sqrt[3]{7} \text{ (real solution)}
\]
The other two solutions will be complex, resulting from the angle rotations in the complex plane:
The cube roots of \(-7\) can be represented in polar form as:
\[
x_k = 7^{1/3} \left( \cos\left(\frac{\pi + 2k\pi}{3}\right) + i \sin\left(\frac{\pi + 2k\pi}{3}\right) \right), \quad k = 0, 1, 2
\]
Thus, the remaining two non-real solutions can be represented as:
- \(x_1 = -\sqrt[3]{7} + \frac{\sqrt[3]{7}}{2} i\sqrt{3}\) (complex)
- \(x_2 = -\sqrt[3]{7} - \frac{\sqrt[3]{7}}{2} i\sqrt{3}\) (complex)
Therefore, for the polynomial \(x^3 + 7 = 0\):
- **Number of real solutions:** 1 (specifically \(-\sqrt[3]{7}\))
- **Number of complex (non-real) solutions:** 2
### Summary
1. For \(x^2 + 5 = 0\):
- Real solutions: 0
- Complex solutions: 2
2. For \(x^3 + 7 = 0\):
- Real solutions: 1
- Complex solutions: 2
### 1. \( x^2 + 5 = 0 \)
Rearranging the equation gives:
\[
x^2 = -5
\]
To find the solutions, we take the square root of both sides:
\[
x = \pm \sqrt{-5} = \pm i\sqrt{5}
\]
Since the solutions involve the imaginary unit \(i\), both solutions are complex. Therefore, we can conclude:
- **Number of real solutions:** 0
- **Number of complex (non-real) solutions:** 2 (specifically \(i\sqrt{5}\) and \(-i\sqrt{5}\))
### 2. \( x^3 + 7 = 0 \)
Rearranging the equation gives:
\[
x^3 = -7
\]
Taking the cube root of both sides gives:
\[
x = \sqrt[3]{-7}
\]
The cube root of a negative number is a real number, and there is exactly one real solution. However, there may be complex solutions as well. The cubic polynomial can be factored:
The three solutions to \(x^3 + 7 = 0\) can be found using polar coordinates or other techniques for cubics. In the complex plane, we know that:
1. One real solution: \(x = -\sqrt[3]{7}\)
2. Two non-real solutions can be found by using the polar form. The other solutions will be in the form:
\[
x = -\sqrt[3]{7} \text{ (real solution)}
\]
The other two solutions will be complex, resulting from the angle rotations in the complex plane:
The cube roots of \(-7\) can be represented in polar form as:
\[
x_k = 7^{1/3} \left( \cos\left(\frac{\pi + 2k\pi}{3}\right) + i \sin\left(\frac{\pi + 2k\pi}{3}\right) \right), \quad k = 0, 1, 2
\]
Thus, the remaining two non-real solutions can be represented as:
- \(x_1 = -\sqrt[3]{7} + \frac{\sqrt[3]{7}}{2} i\sqrt{3}\) (complex)
- \(x_2 = -\sqrt[3]{7} - \frac{\sqrt[3]{7}}{2} i\sqrt{3}\) (complex)
Therefore, for the polynomial \(x^3 + 7 = 0\):
- **Number of real solutions:** 1 (specifically \(-\sqrt[3]{7}\))
- **Number of complex (non-real) solutions:** 2
### Summary
1. For \(x^2 + 5 = 0\):
- Real solutions: 0
- Complex solutions: 2
2. For \(x^3 + 7 = 0\):
- Real solutions: 1
- Complex solutions: 2