If α and β are the zeros of the polynomial px² + qx +r then find the value of α³β + β³α
plz answer!!!!
Answers
Answered by
2
Answer:
Hi buddy
Step-by-step explanation:
GIVEN THAT:)
Zeroes of polynomial px²+qx+r=0
alpha and beta are zeroes of polynomial
then,
a+b=-b/a
a+b=-q/p
Again..
ab=c/a
ab=r/p
then,
value of
a³+b³
(a+b)(a²+b²-ab)
(-q/p){(a+b)²-2ab -ab
(-q/p){(-q/p)²-2*r/p -r/p
(-q/p) [(q²/p²-2r/p -r/p
Next step see in attachment
that's all
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Answered by
2
Answer:
r/p (q²/p² - 2r/p)
Step-by-step explanation:
Given:
- px² + qx +r, α and β are the zeros
Sum of roots:
- α+β= -q/p
Product of roots:
- α*β =r/p
Solution:
- α³β + β³α=
- αβ(α²+β²)=
- αβ((α+β)²-2αβ)=
- r/p((-q/p)²-2r/p)=
- r/p (q²/p² - 2r/p)
Solved
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