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Answer:

To find the zeros of the quadratic equation x² - 4x + 5, we can use the quadratic formula. The quadratic formula states that for any quadratic equation of the form ax² + bx + c = 0, the zeros can be found using the formula:

x = (-b ± √(b² - 4ac))/(2a)

In this case, the coefficients of the quadratic equation are a = 1, b = -4, and c = 5. Plugging these values into the quadratic formula, we get:

x = (-(-4) ± √((-4)² - 4(1)(5)))/(2(1))

x = (4 ± √(16 - 20))/2

x = (4 ± √(-4))/2

Here, we have a square root of a negative number (√(-4)), which means that the zeros will be complex numbers. Let's simplify the equation further:

x = (4 ± 2i)/2

Now, we can simplify the expression by dividing both the numerator and denominator by 2:

x = 2 ± i

So, the zeros of the quadratic equation x² - 4x + 5 are:

x = 2 + i

x = 2 - i

These are the complex zeros of the equation.

Explanation:

User HMcG
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