Final answer:
To prove that the equation x^5+ax³+bx+c=0 has exactly one solution for a>0 and b>0, we can use the fact that the coefficients a and b are positive. By assuming the equation has two solutions and reaching a contradiction, we can conclude that the equation indeed has only one solution.
Step-by-step explanation:
To prove that the equation x5+ax³+bx+c=0 has exactly one solution for a>0 and b>0, we can use the fact that the coefficients a and b are positive. Let's assume that the equation has two solutions, say x = p and x = q. Then, if we substitute these values into the equation, we will get:
p5+ap³+bp+c=0
q5+aq³+bq+c=0
Now, let's subtract the second equation from the first equation:
p5-q5+ap³-aq³+bp-bq=0
This equation can be rewritten as:
(p-q)(p4+p³q+p²q²+pq³+q4)+a(p-q)(p²+pq+q²)+b(p-q)(p+q)+c=0
Since a and b are positive, (p-q)(p²+pq+q²) will always be positive. Additionally, (p-q)(p+q) will also be positive since a and b are positive. Therefore, for the equation to hold true, (p-q)(p4+p³q+p²q²+pq³+q4) must be negative. However, this is not possible since all the terms in the parentheses are non-negative. Therefore, our initial assumption that the equation has two solutions is incorrect. Hence, the equation x5+ax³+bx+c=0 has exactly one solution for a>0 and b>0.