1. while solutions to the general quintic equation in terms of radicals are impossible to obtain, they do exist. prove that the given quintic equation has real solutions. x^5-3x+1=0

Answer with step-by-step explanation:

Quintic function :It is defined as the function of degree five.

It can be defined

Where,b,c,d,e and f are members of field , typically rational numbers,real numbers and complex numbers .

We are given that quintic equation

We have to prove that given quintic equation has real solutions

Using Discarte's rule

When are going from coefficient of first term to coefficient of second term then sign change and when we are going from coefficient of second term to coefficient of third term then sigh also changes .Therefore,, there are two times sign change .Hence, possible positive real roots are 2 or zero.Then number of positive real roots are even in number .

When we are going from first term to second term then sign change and then from second term to third term then sign does not changes.

Hence, possible negative real root is   1.

Therefore, we can say given quintic equation has real root.

Hence, proved

answer:

a=9

step-by-step explanation: lmk if you need a explanation

7/2

(decimal: 3.5)

welcome ; )

step-by-step explanation: