Asked by yeslian c
Given the equation x3+6x2−10x−60=0
, which of the following describes a possible solution set?(1 point)
Responses
The equation has all non-real solutions.
The equation has all non-real solutions.
The equation has four non-real solutions.
The equation has four non-real solutions.
The equation has all real solutions.
The equation has all real solutions.
The equation has two real solutions and one non-real solution.
, which of the following describes a possible solution set?(1 point)
Responses
The equation has all non-real solutions.
The equation has all non-real solutions.
The equation has four non-real solutions.
The equation has four non-real solutions.
The equation has all real solutions.
The equation has all real solutions.
The equation has two real solutions and one non-real solution.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To analyze the equation \( x^3 + 6x^2 - 10x - 60 = 0 \), we can use the following approach to determine the nature of the roots.
1. **Identifying the Degree of the Polynomial**: The equation is a cubic polynomial (degree 3), which means it can have 3 roots. The roots can be real or non-real (complex).
2. **Using the Discriminant**: For cubic equations, the discriminant can help determine the nature of the roots, but it's often easier to analyze this using numerical methods or calculus.
3. **Graphing**: Plotting the function \( f(x) = x^3 + 6x^2 - 10x - 60 \) can illustrate the behavior of the function and where it crosses the x-axis (real roots).
4. **Calculating Values at Specific Points**: We can evaluate \( f(x) \) at various points:
- \( f(-10) = (-10)^3 + 6(-10)^2 - 10(-10) - 60 = -1000 + 600 + 100 - 60 = -360 \)
- \( f(-5) = (-5)^3 + 6(-5)^2 - 10(-5) - 60 = -125 + 150 + 50 - 60 = 15 \)
- \( f(0) = 0^3 + 6(0^2) - 10(0) - 60 = -60 \)
- \( f(5) = 5^3 + 6(5)^2 - 10(5) - 60 = 125 + 150 - 50 - 60 = 165 \)
From these evaluations:
- \( f(-10) < 0 \)
- \( f(-5) > 0 \) (the function changes signs between \( x = -10 \) and \( x = -5 \), indicating a root in that interval)
- \( f(-5) > 0 \) (and continues to \( f(0) < 0 \), so there's another root between \( -5 \) and \( 0 \))
- \( f(0) < 0 \) and \( f(5) > 0 \) (indicating a third root between \( 0 \) and \( 5 \))
Thus, we can conclude that there are three real roots based on the changes in signs between the evaluated points.
5. **Conclusion**: The original cubic equation has three real roots. Therefore, the correct conclusion about the nature of the solutions is:
**The equation has all real solutions.**
1. **Identifying the Degree of the Polynomial**: The equation is a cubic polynomial (degree 3), which means it can have 3 roots. The roots can be real or non-real (complex).
2. **Using the Discriminant**: For cubic equations, the discriminant can help determine the nature of the roots, but it's often easier to analyze this using numerical methods or calculus.
3. **Graphing**: Plotting the function \( f(x) = x^3 + 6x^2 - 10x - 60 \) can illustrate the behavior of the function and where it crosses the x-axis (real roots).
4. **Calculating Values at Specific Points**: We can evaluate \( f(x) \) at various points:
- \( f(-10) = (-10)^3 + 6(-10)^2 - 10(-10) - 60 = -1000 + 600 + 100 - 60 = -360 \)
- \( f(-5) = (-5)^3 + 6(-5)^2 - 10(-5) - 60 = -125 + 150 + 50 - 60 = 15 \)
- \( f(0) = 0^3 + 6(0^2) - 10(0) - 60 = -60 \)
- \( f(5) = 5^3 + 6(5)^2 - 10(5) - 60 = 125 + 150 - 50 - 60 = 165 \)
From these evaluations:
- \( f(-10) < 0 \)
- \( f(-5) > 0 \) (the function changes signs between \( x = -10 \) and \( x = -5 \), indicating a root in that interval)
- \( f(-5) > 0 \) (and continues to \( f(0) < 0 \), so there's another root between \( -5 \) and \( 0 \))
- \( f(0) < 0 \) and \( f(5) > 0 \) (indicating a third root between \( 0 \) and \( 5 \))
Thus, we can conclude that there are three real roots based on the changes in signs between the evaluated points.
5. **Conclusion**: The original cubic equation has three real roots. Therefore, the correct conclusion about the nature of the solutions is:
**The equation has all real solutions.**
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