Question
Which system of equations has no real solutions?
answer choices
y = x^2 and y = x + 1
y = x^2 + 1 and y = -x^2 - 1
y = -x^2 + 2 and y = x^2
y = 3x + 1 and y = 2x + 2
answer choices
y = x^2 and y = x + 1
y = x^2 + 1 and y = -x^2 - 1
y = -x^2 + 2 and y = x^2
y = 3x + 1 and y = 2x + 2
Answers
GPT-4o mini
To determine which system of equations has no real solutions, we can analyze each pair of equations by looking for intersections or the possibility of having points in common.
1. **y = x^2 and y = x + 1**
- The first equation is a parabola opening upwards, and the second is a straight line. Setting \(x^2 = x + 1\) gives a quadratic equation \(x^2 - x - 1 = 0\), which has real solutions (the discriminant is positive).
2. **y = x^2 + 1 and y = -x^2 - 1**
- The first equation is a parabola shifted up by 1, and the second equation is a parabola opening down, shifted down by 1. Setting \(x^2 + 1 = -x^2 - 1\) gives \(2x^2 + 2 = 0\) or \(x^2 = -1\), which has no real solutions since the square of a real number cannot be negative.
3. **y = -x^2 + 2 and y = x^2**
- The first equation is a downward-opening parabola, and the second is an upward-opening parabola. Setting \(-x^2 + 2 = x^2\) gives \(2 = 2x^2\) or \(x^2 = 1\), which has real solutions (\(x = 1\) and \(x = -1\)).
4. **y = 3x + 1 and y = 2x + 2**
- Both equations are lines. Setting \(3x + 1 = 2x + 2\) gives \(x = 1\), which means they intersect at the point \((1, 4)\). So, there are real solutions.
From the analysis above, the system of equations that has no real solutions is:
**y = x^2 + 1 and y = -x^2 - 1**.
1. **y = x^2 and y = x + 1**
- The first equation is a parabola opening upwards, and the second is a straight line. Setting \(x^2 = x + 1\) gives a quadratic equation \(x^2 - x - 1 = 0\), which has real solutions (the discriminant is positive).
2. **y = x^2 + 1 and y = -x^2 - 1**
- The first equation is a parabola shifted up by 1, and the second equation is a parabola opening down, shifted down by 1. Setting \(x^2 + 1 = -x^2 - 1\) gives \(2x^2 + 2 = 0\) or \(x^2 = -1\), which has no real solutions since the square of a real number cannot be negative.
3. **y = -x^2 + 2 and y = x^2**
- The first equation is a downward-opening parabola, and the second is an upward-opening parabola. Setting \(-x^2 + 2 = x^2\) gives \(2 = 2x^2\) or \(x^2 = 1\), which has real solutions (\(x = 1\) and \(x = -1\)).
4. **y = 3x + 1 and y = 2x + 2**
- Both equations are lines. Setting \(3x + 1 = 2x + 2\) gives \(x = 1\), which means they intersect at the point \((1, 4)\). So, there are real solutions.
From the analysis above, the system of equations that has no real solutions is:
**y = x^2 + 1 and y = -x^2 - 1**.