Thread: Another problem from a USA contest

1. Another problem from a USA contest

Thank you,I will try to do it myself.Can someone now help me with this one:
1983.).Prove that has at least one real solution if

2. Originally Posted by myro111
Thank you,I will try to do it myself.Can someone now help me with this one:
1983.).Prove that the solutions of are real if
There is obviously something missing here. The given condition only uses the coefficients a and b. But if you increase the constant term e sufficiently, you can always ensure that the equation only has one real solution.

3. I edited the post,I wrote the problem wrong,my mistake,sry.

4. if $2*a^2<5*b$ then we know that b must be positive. There is no way be can be negative and still be greater than $2*a^2$. a is either positive or negative, so we have 2 cases:

a is positive: if a is positive, we apply the rule of signs. Notice that the first coefficient of the polynomial is 1. Once we do f(-x) to check for negative roots, we see that there is a sign change: from the first term to the second term. This shows that there is at least one real negative root.

a is negative: if a is negative, we apply the rule of signs once more. Since b must be positive, we see a sign change from the second to third term, and this proves the existence of at least one real positive root.

If you are unfamiliar with Descartes' Rule of Signs, take a look at this:

Descartes' Rule of Signs

5. Originally Posted by myro111
Thank you,I will try to do it myself.Can someone now help me with this one:
1983.).Prove that has at least one real solution if
An odd order polynomial equation with real coefficients always has at least one real root (if you need a proof the intermediate value theorem will do coupled with the result that for any odd order polynomial p(x) with real coefficients for x large enough p(x) and p(-x) are of opposite signs)

CB

6. Originally Posted by rtblue
if $2*a^2<5*b$ then we know that b must be positive. There is no way be can be negative and still be greater than $2*a^2$. a is either positive or negative, so we have 2 cases:

a is positive: if a is positive, we apply the rule of signs. Notice that the first coefficient of the polynomial is 1. Once we do f(-x) to check for negative roots, we see that there is a sign change: from the first term to the second term. This shows that there is at least one real negative root.

a is negative: if a is negative, we apply the rule of signs once more. Since b must be positive, we see a sign change from the second to third term, and this proves the existence of at least one real positive root.
No it does not, you need to show an odd number of sign changes to guarantee at least one root. If there are n sign changes then there are n, n-2, n-4 .. (ending at either 1 or 0 depending on the parity of n) roots of the appropriate sign. So you can guarantee that there is at least one positive root only if p(x) has an odd number of sign changes in the coefficients or that p(-x) has an odd number of sign changes.

CB

7. I apologize for that oversight, Captain Black. I forgot that the number of roots could be n,n-2,... Thank you.