A polynomial on it's own does not have any solutions.
ax2 + bx + c
has no solutions, since it is not an equation. When we say the roots or solutions of ax2 + bx + c we are usually assuming it equals 0 and is therefore an equation.
For ALL real values of x, an nth degree polynomial (one which starts axn + ...) has n solutions when equal to 0. Some of them might be the same, for example:
x4 - 6x3 + 13x2 - 12x + 4 = 0
This has four solutions: x = 1, 1, 2 or 2. Yes, obviously some of them are the same, but it doesn't mean they're not there. This is what you'd expect from a polynomial of degree 4.
Interestingly enough, if the polynomial is of degree 1/n, it will still have n solutions (figure it out).
Geometrically speaking:
the polynomial ax2 + bx + c = 0 always has 2 solutions. If they are different, then the line y = 0 crosses the parabola at two places. If the solutions are both the same then the line y = d is a tangent to the parabola. If the solutions are complex (i.e. if b2 - 4ac is negative), the lines do not cross.
With the example above, two pairs of identical solutions means that the polynomial above is a tangent to the line y = 0 at 2 places , (1, 0) and (2, 0).
So to finish:
1) Any equation where xn is the highest power of x, has n roots (or 1/n roots if it's x1/n).
2) If the roots form in pairs, then geometrically you have a tangent.
3) So let's stop arguing about y = x0.5, and get on with our lives
OK although there is some truth in that, I wrote it at gone midnight and the tiredness certainly shows. A fair amount of that is bollocks and I only realised this after I went to bed so let's leave this as an example of why not to post when you're tired.
everything you said is almost right. EXCEPT, first of all
(4)^(1/2) is a NUMERICAL VALUE, not two, and 4 itself is NOT A VARIABLE (don't tell me that 4^3+3*4^2+2*4+1=0 has 3 solutions!!)
and also, about the n solutions thing. no one says all n solutions are real solutions! there are infact a good chance of been complex! thats why there is only one REAL solution to x^(1/2)=1, when x=1. the other solution to it is complex. (hey don't tell me x=-1 is a solution to that problem)
This is one of my first posts so don't start on me yet...
From a professional point of view (I lecture math at Exeter University, UK) there is one correction that immediately comes to mind from Gordon's original post:
1) A polynomial of nth degree when equated to 0 has n real solutions if n is even, and 1 real solution if n is odd. Same applies to the 1/n thing. To be fair Gordon didn't say he/she didn't know about the complex roots of an equation, since he did only specify the real part of the story.
Complex solutions always appear in [edit]conjugate[/edit] pairs, and the complex solutions of a polynomial to the 3rd degree, which only cuts the x axis once has two complex solutions and one real one... I think that's what he meant.
Originally posted by Masterbandit666 This is one of my first posts so don't start on me yet...
From a professional point of view (I lecture math at Exeter University, UK) there is one correction that immediately comes to mind from Gordon's original post:
1) A polynomial of nth degree when equated to 0 has n real solutions if n is even, and 1 real solution if n is odd. Same applies to the 1/n thing. To be fair Gordon didn't say he/she didn't know about the complex roots of an equation, since he did only specify the real part of the story.
So, similar to bugz's point of odd, couldn't a 4th order polynomial have 2 complex and 2 real solutions, or even just 4 complex solutions? Or, more generically stated:
A polynomial of nth degree when equated to 0 has, if n is even,
{2*a} real solutions and {2*b} complex solutions
where {2*a + 2*b = n; and a, b are non-negative}.
It took me a while to come up with this, but maybe masterwotsit meant that x^n = 0 has 1 root when n is odd... that fits his definition but that's not what he said. Are you sure you lecture math?
Spiderman was $h1t
MIB2 was $h1t
Goldmember was $h1t
So someone name a recent film that was hyped up to be great but actually WAS...
Someone correct me if I'm wrong, but I don't beleive it's possible to have an equation with 6 roots and have them all real. It is my understanding that 5 is the limit.
x has 6 real roots, 1, -2, 3, -4, 5, and -6, or unless I'm completely misunderstood, but on another note, if 5 is the limit, and you have x^8, 5 MUST be real, but the other 3 can't be complex because there is always an even number of complex roots.
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