 # if p and q are the roots of the equation ax^2+bx+c=0,find the value of p^5+q^5.plz show the steps. 10 years ago

p^5 + q^5 = (p+q)*{ P^4 - (P^3)*q + (P^2)*(q^2) - p*(q^3) + q^4 }

Now, p^4+q^4 = {p^2 + q^2}^2 - 2*(P^2)*(q^2)

similarly p^2+q^2 = (p+q)^2 - 2*(pq)

hence you have pow(p,4) + pow(q,4) interms of p+q and pq, the sum and product of roots.

for the oter terms in the flower brackets in (1) take (p*q) common, and you will get that interms of (p+q) and (pq)....

Hope that helps!!!! (I have just mentioned one of the approaches of how to solve, you need to solve it completely) :-)

10 years ago

(a+b)^5=(a^5)+(b^5)+10((ab)^2)(a+b)+5ab(a^3+b^3)

==>(a+b)^5=(a^5)+(b^5)+10((ab)^2)(a+b)+5ab[((a+b)^3)-3ab(a+b)]

==>(a^5)+(b^5)=[(a+b)^5]-10((ab)^2)(a+b)-5ab[((a+b)^3)-3ab(a+b)]................(expansion of a^5+b^5)...............(1)

substituting (a+b) as (-b)/a...............(sum of the roots)

and ab as c/a.................(porduct of the roots)

in the equation 1 and simplifying it we get [(-b/a)^5]+[5(b^3)c]/(a^4)-[5b(c^2)]/(a^3)

therefore p^5+q^5=[(-b/a)^5]+[5(b^3)c]/(a^4)-[5b(c^2)]/(a^3)......