Not for all types of equations. But always in second degree equations they do.
Consider a third degree equation with 3 different roots. Obviously, one of the roots can not be in a pair.
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In the complex field, the two numbers are the same. If you restrict yourself to real solutions, the relationship is as follows: A polynomial of degree p has p-2k real solutions where k is an integer such that p-2k is non-negative. [There will be 2k pairs of complex conjugate roots.]
Oh, isn't that just a happy little question? The complex conjugate of a real number like 8 is just 8 itself because there is no imaginary part to change. Just like how every tree needs its roots, every real number needs its complex conjugate to stay balanced and harmonious. Just remember, there are no mistakes, only happy little accidents in math!
There are 3 cube roots and these are:the real root -1.2599and the complex roots 0.6300 - 1.0911i and its conjugate, 0.6300 + 1.0911i.
It can be used as a convenient shortcut to calculate the absolute value of the square of a complex number. Just multiply the number by its complex conjugate.I believe it has other uses as well.
The equation is:x3 - 4x2i - 9xi2+ 36i3 = 0Note: The ' i ' must be an ordinary constant, and can't be sqrt(-1).If it were sqrt(-1), then -4i would also be a root of the equation.Imaginary or complex roots always occur in conjugate pairs.If -4i is also a root of the equation and the questioner just forgot to include it,then ' i ' can be sqrt(-1), and the equation can bex4 + 25x2 + 144 = 0