The equation y=x2−4x+5

Question

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

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
two real and zero non-real

zero real and zero non-real

Answer 1

To determine the number of real and non-real solutions for the quadratic equation \( y = x^2 - 4x + 5 \), we can analyze its discriminant. The discriminant \( D \) for a quadratic equation in the form \( ax^2 + bx + c \) is given by:

\[ D = b^2 - 4ac \]

In our equation:

• \( a = 1 \)
• \( b = -4 \)
• \( c = 5 \)

Now we can calculate the discriminant:

\[ D = (-4)^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4 \]

Since the discriminant \( D \) is negative, it indicates that there are no real solutions. Instead, there are two complex (non-real) solutions.

Thus, the correct response is:

zero real and two non-real.