Question

complex numbers review
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the vertex of the parabola is at (2, 1). the approximate coordinates of the curve are as follows: (-1, 10), (0, 5), (2, 1), (4, 5), and (5, 10).
the equation ( y = x^2 - 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)
○ zero real and two non - real
○ two real and zero non - real
○ one real and one non - real
○ zero real and zero non - real

Explanation:

Step1: Analyze the parabola's graph

The vertex of the parabola is at \((2, 1)\), and all the given points on the curve have \(y\)-values greater than or equal to \(1\) (e.g., \((-1, 10)\), \((0, 5)\), \((2, 1)\), \((4, 5)\), \((5, 10)\)). This means the parabola opens upwards (since the coefficient of \(x^2\) in \(y = x^2 - 4x + 5\) is positive) and its minimum \(y\)-value is \(1\) (at the vertex).

Step2: Relate to real solutions

A real solution of the equation \(y = x^2 - 4x + 5\) (or \(x^2 - 4x + 5 = 0\) when \(y = 0\)) corresponds to the \(x\)-intercepts of the parabola (where the graph crosses the \(x\)-axis, i.e., \(y = 0\)). Since the minimum \(y\)-value of the parabola is \(1\), which is greater than \(0\), the parabola never crosses the \(x\)-axis. Therefore, the equation \(x^2 - 4x + 5 = 0\) has no real solutions.

Step3: Determine non - real solutions

For a quadratic equation \(ax^2+bx + c = 0\) (\(a
eq0\)), the discriminant is \(\Delta=b^2 - 4ac\). For the equation \(x^2 - 4x + 5 = 0\), \(a = 1\), \(b=-4\), and \(c = 5\). The discriminant \(\Delta=(-4)^2-4\times1\times5=16 - 20=-4\). Since the discriminant is negative, the quadratic equation has two non - real (complex) solutions. Also, from the fundamental theorem of algebra, a quadratic equation has exactly two solutions (counting multiplicities) in the complex number system. Since there are no real solutions, there must be two non - real solutions.

Answer:

zero real and two non - real (the option: zero real and two non - real)