That is true if the polynomial has real coefficientes, an this is not the case. Substitute and you will find .
Fernando Revilla
Hey guys I'm not sure whether my approach to this question is right...kinda looks simple enough eh?
QUESTION
One root of is . Find the other root of this equation and the value of K in the form
So this is what I did:
Since is a root, is also a root as complex roots come in conjugate pairs
But then?
have I misunderstood something?...im rusty with complex numbers
The actual answers are that the other root is ,and
Thanks
That is true if the polynomial has real coefficientes, an this is not the case. Substitute and you will find .
Fernando Revilla
If a polynomial has REAL coefficients, then what you said was correct. That is, roots come in complex conjugate pairs. However, this polynomial has a complex coefficient, so what you said doesn't hold (sorry!).
Personally, I would just work out , using the fact that the coefficient of must be . It is, perhaps, crude. But it works.
infact you must ontain the roots of a quadratic function:
Quadratic function - Wikipedia, the free encyclopedia
in this case:
a=1, b=1-i, c=-13i
X1,2=(-b+-sqrt(b^2-4ac))/2a
This is all a bit too much though - as the OP pointed out, to do it this way you need to find the square root of a complex number, and so we are spiralling down a route that, although correct, requires a lot of work.
It is much easier to approach the question in the way I pointed out in my post.
So, what are a and b? So what is the other root, and what is K?
Simples...