Question

Prove that the equation:
ax^3+bx^2+cx+d=0 has atleast one real root.

Answer 1

Let f(x) ax^3 + bx^2 + cx + d.
This function, when graphed, will look like a cubic equation.
Have you learned about end behavior of functions? (If not, I think I can explain it.)
If so, well, for some very large number m, the value of the function will be some very large positive number.
And for some very large negative number n, the value of the function will be some very large negative number. (Or maybe the other way around.)
So in between, there must be a point where the graph crosses the x-axis. So thus there is at least one real root.
Does that make sense?

Answer 2

@zonazoo

Answer 3

what is m though? is that what u are substituting for a? what if its a small number?

Answer 4

m is just some large number that I chose because f(m) is large. So is n.
We want to make m a large number so that f(m) is large, or greater than 0. Here's a picture:

Answer 5

See, if m's small, it might not work. If m's large, however, then we can see that it becomes a large positive number eventually.
Does that make more sense?

Answer 6

well looking at the graph when you move it up or down you can see that it will cross atleast once... but is that some rule you use to prove this, or do i just have to say for some large m and large -n

Answer 7

I used two theorems: the intermediate value theorem (I think that's what it's called) which is usually taught in Calculus, I don't know why not earlier, and the polynomial end behavior theorem.
The p.e.b. theorem says that if you have a polynomial function of odd degree, than it goes "up" at one end and "down" at the other.
The i.v.t. says that if you have a function with no breaks or anything, and there's one value that's positive, and there's another value a little way off that's negative, there's got to be a root somewhere in between.
Does that make sense? Can you see where I used the theorems?

Answer 8

Thanks for the medal. :) I wish I could give you two medals. This was, like, my favorite problem on OpenStudy to solve, ever.

Answer 9

thank you for your help. :-)