Question

If p, q, r are the real roots of x 3 − 6x2 + 3x + 1 = 0, determine the possible values of p 2 q + q 2 r + r 2 p.

Answer 1

Answer:

Step-by-step explanatio

We know that (pq+qr+rp)=c/a

Here c=3 and a=1,

Therefore the value of

= p2q+q2r+r2p

=2(pq+qr+rp)

=2×3=6.

Answer 2

Answer:

6

Step-by-step explanation:

Here in the equation x 3 − 6x2 + 3x + 1 = 0

a= 1, b=-6 , c= 3 and d= 1

since, p,q,r are are roots of the equation we have

pq+rq+pr = c/a

⇒ pq+rq+pr = 3/1

Now,  p 2 q + q 2 r + r 2 p =2(pq+rq+pr)  = 2×3 =6