why is it not possible for a quadratic equation to have one real root and one imaginary root? what would be one example that demonstrates this? 

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Suppose the quadratic has one real root. Recall that the graph is symmetric about a vertical line (the axis of symmetry) so every point save one has a mirror image point and thus two real zeros.

The exceptional case is when the zero is the vertex of the graph. This...

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Suppose the quadratic has one real root. Recall that the graph is symmetric about a vertical line (the axis of symmetry) so every point save one has a mirror image point and thus two real zeros.

The exceptional case is when the zero is the vertex of the graph. This point is called a double root. The discriminant of the function will be zero and the function will factor as a binomial squared, again having two real roots. (Both are the same.)

Imaginary roots must come in pairs. We are again adding and subtracting from the axis of symmetry, but now we are adding/subtracting an imaginary number.

Consider `y=x^2+2x+2` . To find the zeros:

`x^2+2x+2=0`

`x^2+2x=-2`

`x^2+2x+1=-2+1`

`(x+1)^2=-1`

`x=-1+-i`  In taking the square root, we must keep both possible answers getting an imaginary number and its conjugate.

If the right side was zero the solution would be real, and if the right side was positive there would be two real solutions.

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