Final answer:
The systems y = x^2 + 4x + 7 and y = 2, and y = –x^2 – 3 and y = 9 + 2x have no real number solutions as their corresponding quadratic equations have a negative discriminant.
Step-by-step explanation:
To determine which systems of equations have no real number solutions, we need to analyze each pair of equations to see if their graphs intersect. Systems that do not intersect have no real number solutions because the equations represent lines or curves that never meet.
- y = x2 + 4x + 7 and y = 2
- y = x2 – 2 and y = x + 5
- y = –x2 – 3 and y = 9 + 2x
- y = – 3x – 6 and y = 2x2 – 7x
- y = x2 and y = 10 – 8x
By setting the quadratic and linear equations equal to each other and attempting to solve for x, we can check if there are real solutions. If the resulting quadratic equation has a negative discriminant (where the discriminant is b2 – 4ac from the quadratic formula), then there are no real solutions. Let's analyze the given pairs:
- For the first pair, setting the equations equal to each other gives x2 + 4x + 5 = 0. The discriminant is 42 – 4(1)(5) = 16 – 20 = -4, which is negative. So, this system has no real number solutions.
- The second pair results in x2 – x – 7 = 0. The discriminant is (-1)2 – 4(1)(-7) = 1 + 28 = 29, which is positive, indicating two real solutions.
- In the third pair, –x2 – 3 = 9 + 2x, or –x2 – 2x – 12 = 0. The discriminant is (-2)2 – 4(-1)(-12) = 4 – 48 = -44, negative, leading to no real solutions.
- The fourth pair, when set equal, yields – 3x – 6 = 2x2 – 7x, or 2x2 – 4x – 6 = 0. The discriminant is (-4)2 – 4(2)(-6) = 16 + 48 = 64, which is positive, so there are real solutions here.
- Finally, for the fifth pair, x2 = 10 – 8x, or x2 + 8x – 10 = 0. The discriminant is (8)2 – 4(1)(-10) = 64 + 40 = 104, which is positive, indicating real solutions.
Therefore, the systems y = x2 + 4x + 7 and y = 2, and y = –x2 – 3 and y = 9 + 2x have no real number solutions.