Question

If p,q,r are the zeroes of the polynomial x³+4x+1, then find the value of 1/p+q +1/q+r + 1/p+r..

Answer 1

Answer:

Step-by-step explanation:

x³+4x+1=0

As p,q, r are zeroes of the above polynomial, we have

p+q+r=0; pq+qr+pr=4; pqr= -1 ...(I)

(Using alternate value theorem i.e. if ax³+bx²+cx+d=0 is a polynomial, then

Sum of the product of the roots taken 1 at a time= -b/a=p+q+r

Sum of the product of the roots taken 2 at a time= c/a=pq+qr+pr

Sum of the product of the roots taken 3 at a time= d/a=pqr)

Now, 1/(p+q)+1/(q+r)+1/r+p)

= - (1/p + 1/q +1/r)  [As p+q= -r, q+r= -p, p+r=-q...from (I)]

= - (pq+qr+pr)/pqr

= - (4)/(-1) ...from (I)]

= 4 (Answer)