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# 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. Grade:12

## 2 Answers

9 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)....

Substitute from the quadratic eqn form sum and prdt of roots, and you have your answer.....

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

9 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)......

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