I totally got confused by real roots so are complex roots like fake roots but what does that mean?

Question

THIS IS WHAT I GOT FOR ONE OF MY PREVIOUS QUESTIONS:
How do I show that the equation x^5 +2x^3 -x^2 -2x -6 only has one solution without using the graphical method?
First- Any nth degree (here n=5) polynomial has at most n complex roots.

Second : One way would be to find all the roots
root1 = 1.395917950,
root2 = .1362911182+1.710032257*I,
root3 = -.8342500933+.8744321762*I,
root4 = -.8342500933-.8744321762*I, .
root5 = 1362911182-1.710032257*I
ie 4 complex roots, one real root. All 5 roots have been found and only one is real.

but i don't understand what the other complex roots mean. like there is fake roots that don't cut the x-axis so where do these complex roots cut? Please explain anything that will help me understand

Answer 1

Do a google search on "Descartes Rule of Signs" and you will learn all you need to know.

Also, the complex roots come in matched pairs known as "complex conjugates." To make up an example, suppose your roots are x = 3 + 2i and x = 3 - 2i.

Then x - (3 + 2i) = 0 and x - (3 - 2i) = 0.

[x - (3 + 2i)] [x - (3 - 2i)] = 0

[(x-3) + 2i] [(x-3) - 2i] = 0

(x-3)^2 - (2i)^2 = 0

(x^2 - 6x + 9) - 4i^2 = 0

(x^2 - 6x + 9) - (-4) = 0

x^2 - 6x + 13 = 0

This quadratic has no real roots. Its pair of roots are called "complex conjugates." If you graph this function y = f(x), you'll see that it never crosses the x-axis. To get the minimum, take the derivative:

dy/dx = 2x - 6 = 0 ==> x = 3

f(3) = 3^2 - 6 x 3 + 13 = 9 - 18 + 13 = 4 > 0

You can solve by the quadratic formula or by completing the square:

x^2 - 6x + 9 = -13 + 9 = -4
(x - 3)^2 = +/- sqrt(-4) = +/- 2i
x = 3 +/- 2i

This solution may be represented as two points in the complex plane, (3, 2i) and (3, -2i), or as vectors 3i +/- i2j. The magnitude of these vectors is sqrt(3^2 + 2^2) = sqrt(13). Therefore, they can be expressed as complex numbers

sqrt(13)(sin theta +/- i cos theta)

where the angle theta is determined by tan theta = 2/3.

Finally, to directly answer your question, "are complex roots like fake roots, but what does that mean?", I'd say that the complex roots are roots that satisfy the original equation. If you plug in the complex roots, the original equation is satisfied.

A complex number, such as 3 + 2i, typically has a real portion and an imaginary portion. The imaginary portion is not a real number, and it involves the square root of negative one. Nevertheless, complex numbers can be graphed in the complex plane, and complex numbers have properties that make them useful in mathematics and, for example, in the theory of electrical circuits. If you study transient effects in electromagnetics, you learn the value of complex numbers in a big hurry.

And one final thought about graphing. You can graph a complex function in three dimensions, with x, y, and z coordinates. Let x be the real portion, y the imaginary portion, and z the magnitude, i.e., sqrt(x^2 + y^2). When you do that, you'll see that these complex solutions cut some of the axes.

Answer 2

I truthfully have consistently felt that way. truthfully, I truthfully have on no account incredibly bashed absolutely everyone for rooting for a undeniable group, and that i on no account reported something undesirable a pair of group until the celtics followers initiate speaking crap on the subject of the bulls throughout the time of their sequence. I on no account cherished the celtics, yet i on no account published something undesirable on the subject of the gang or the followers until a pair of month in the past. specific i'm getting incredibly nervous while the bulls play in a interest 7! while they performed the celtics in interest 7 i become pacing back and fourth and could not sit down!