![]() Looking for the X values, that would make thisĮxpression equal to 0. So I encourage you to pause this video and try to figure out what Or another way of thinking about it, there's exactly 2 values for X that will make F of X equal 0. Is a second degree polynomial we are going to have exactly 2 roots. The fundamental theorem of algebra tells us that because this Second degree polynomial, 5 X squared plus 6 X plus 5. We have the function F of X being defined by the Hopefully this last bit wasn't confusing, it's just a way to pretend we have the complex numbers on the x,y graph So, since on a complex plane its the vertical axis that has the imaginary numbers, to visualize it on our normal graph, we take the roots of this flipped parabola, make a perfect circle with the points of the two roots on the opposite sides of this circle, then rotate the points 90 degrees, as shown on this website: If this flipped parabola has the equation y= -(x-a)^2+b, then it has the real roots a +/- sqrt(b) whereas our original parabola had the equation y- (x-a)^2+b with the complex roots a +/- sqrt(b)*i. This reflected parabola now has to cross the x=axis, and have 2 real roots. But if we make a vertical axis of the units of i (imaginary numbers), and make the real numbers run along a horizontal axis, the complex root can be seen graphed on this "complex plane." (And all sorts of fun can be had with graphing on the complex plane in precalculus)Īlso, even though the complex roots don't exist on the x,y graph (since there's no such thing as a square root of a negative real number, so there's no way we can plot it on the cartesian plane of real numbers, that's why we need the imaginary plane) it can still be visualized on the x,y graph using a little trick, which is, to take our parabola and flip it upside down. ![]() No, the complex root is not defined by crossing anything, and no, on our regular x,y graph it can't be seen. ![]()
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