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
if then we know that b must be positive. There is no way be can be negative and still be greater than . 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