The equation y=x2−4x+5 has two solutions. Compare this information with the graph of the equation. How many real and non-real solutions does the equation have? (1 point) Responses zero real and zero non-real zero real and zero non-real one real and one non-real one real and one non-real zero real and two non-real zero real and two non-real two real and zero non-real
To find the number of real solutions, we can use the discriminant of the quadratic equation. The discriminant is given by b^2 - 4ac, where a, b, and c are the coefficients in the quadratic equation ax^2 + bx + c = 0.
In this case, the equation y = x^2 - 4x + 5 can be rewritten as x^2 - 4x + (5 - y) = 0. Comparing this to the standard quadratic equation form, we have a = 1, b = -4, and c = 5 - y.
Therefore, the discriminant is given by (-4)^2 - 4(1)(5 - y) = 16 - 20 + 4y = 4y - 4.
For the equation to have real solutions, the discriminant must be greater than or equal to zero. Therefore, we have:
4y - 4 ≥ 0
4y ≥ 4
y ≥ 1
This means that there are real solutions only when y is greater than or equal to 1.
To determine the number of real solutions, we can look at the graph of the equation. A quadratic equation of the form y = ax^2 + bx + c is a parabola. We can see that the coefficient of x^2 (a) is positive, indicating that the parabola opens upwards.
Since the discriminant is greater than or equal to zero when y is greater than or equal to 1, this means that the entire parabola lies above the x-axis for y greater than or equal to 1. Therefore, it has two real solutions.
In summary, the equation y = x^2 - 4x + 5 has two real solutions and zero non-real solutions. The answer is: two real solutions and zero non-real solutions.