if alpha and beta are the roots of ax^2+bx+c then find (alpha)^5+(beta)^5
Answers
Answer :-
α^5 + β^5 = ( - b^5 + 5ab³c - 5a²bc² )/a^5
Solution :-
ax² + bx + c
α, β are zeroes of the polynomial
- Sum of zeroes = α + β = - b/a
- Product of zeroes = αβ = c/a
⇒ (α² + β²)(α³ + β³) = α²(α³ + β³) + β²(α³ + β³)
⇒ (α² + β²)(α³ + β³) = a^5 + α²β³ + α³β² + β^5
⇒ (α² + β²)(α³ + β³) = α^5 + β^5 + α²β²(α + β)
⇒ α^5 + β^5 = (α² + β²)(α³ + β³) - (αβ)²(α + β)
Using identity α² + β² = (α + β)² - 2αβ and α³ + β³ = (α + β)³ - 3αβ(α + β)
⇒ α^5 + β^5 = { (α + β)² - 2αβ }{ (α + β)³ - 3αβ(α + β) } - (αβ)²(α + β)
Substituting the given values
⇒ α^5 + β^5 = { ( - b/a )² - 2( c/a ) }{ ( - b/a )³ - 3( c/a )( - b/a ) } - ( c/a )²( - b/a )
⇒ α^5 + β^5 = (b²/a² - 2c/a)( - b³/a³ + 3bc/a² ) + bc²/a³
⇒ α^5 + β^5 = { (b² - 2ac)/a² }{ ( - b³ + 3abc)/a³ } + bc²/a³
⇒ α^5 + β^5 = { (b² - 2ac)( - b³ + 3abc ) } / a^5 + bc²/a³
⇒ α^5 + β^5 = { b²( - b³ + 3abc ) - 2ac(- b³ + 3abc ) } / a^5 + bc²/a³
⇒ α^5 + β^5 = ( - b^5 + 3ab³c + 2ab³c - 6a²bc² ) / a^5 + bc²/a³
⇒ α^5 + β^5 = ( - b^5 + 5ab³c - 6a²bc² + a²bc² )/a^5
⇒ α^5 + β^5 = ( - b^5 + 5ab³c - 5a²bc² )/a^5