No Real Solutions Quick Check

Question

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Question
Which system of equations has no real solutions?(1 point)
Responses

y = x2+1 and y =−x2 −1
y equals x squared plus 1 and y equals negative x squared minus 1

y=3x+1 and y=2x+2
y equals 3 x plus 1 and y equals 2 x plus 2

y = −x2+2 and y=x2
y equals negative x squared plus 2 and y equals x squared

y= x2 and y= x +1

Answer 1

To determine which system of equations has no real solutions, we need to analyze each pair of equations to see if they intersect or not.

1. y = x² + 1 and y = -x² - 1

   • The first equation (y = x² + 1) is a parabola that opens upwards with a vertex at (0, 1).
   • The second equation (y = -x² - 1) is a parabola that opens downwards with a vertex at (0, -1).
   • Since the vertex of the first parabola is above the x-axis, and the vertex of the second parabola is below the x-axis, these two curves will not intersect.
     No real solutions.

2. y = 3x + 1 and y = 2x + 2

   • These are both linear equations. The first equation has a slope of 3 and the second one has a slope of 2. Since their slopes are different, they will intersect at exactly one point.
     Has real solutions.

3. y = -x² + 2 and y = x²

   • Both are parabolas. The first one opens downward and the second one opens upward. Setting -x² + 2 = x² leads to -2x² + 2 = 0, which means x² = 1, thus x = ±1. So, they intersect at two points.
     Has real solutions.

4. y = x² and y = x + 1

   • The first equation is a parabola opening upwards, while the second equation is a line. Setting x² = x + 1 leads to x² - x - 1 = 0, which has real solutions (by the quadratic formula: x = (1 ± √5) / 2). So they intersect at two points.
     Has real solutions.

In summary, the only system of equations without real solutions is: y = x² + 1 and y = -x² - 1.