If α and β are the zeros of the polynomial f(x)=5x^2+4x−9 then evaluate the following: alpha^3+beta^3​

2 thoughts on “If α and β are the zeros of the polynomial f(x)=5x^2+4x−9 then evaluate the following: alpha^3+beta^3​”

1. Given:–

   • α and β are the zeros of the polynomial [tex]f(x)=5x²+4x−9.[/tex]

   To Find:–

   • Value of α³ + β³

   Formula used:-

   • (a + b)² = a² + b² + 2ab
   • (a – b)² = a² + b² – 2ab
   • a³ + b³ = (a + b) (a² – ab + b²)

   Solution:–

   Comparing the given equation with standard equation i.e. ax² + bx + c = 0 we get,

   • a = 5
   • b = 4
   • c = -9

   Sum of zeroes = [tex]\dfrac{-b}{a}[/tex]

   [tex]⟹α + β = \dfrac{-4}{5}[/tex]

   Product of zeroes = [tex]\dfrac{c}{a}[/tex]

   [tex]⟹αβ = \dfrac{-9}{5}[/tex]

   Using identity, (a + b)² = a² + b² + 2ab

   [tex]⟹(α + β)² = α² + β² + 2αβ[/tex]

   Putting values,

   [tex]⟹(\dfrac{-4}{5})² = α² + β² + 2×\dfrac{-9}{5}[/tex]

   [tex]⟹\dfrac{16}{25}= α² + β² – \dfrac{18}{5}[/tex]

   [tex]⟹α² + β² = \dfrac{16}{25}- \dfrac{18}{5}[/tex]

   [tex]⟹α² + β² = \dfrac{16-90}{25}[/tex]

   [tex]{\boxed{⟹α² + β² = \dfrac{-74}{25}}}[/tex]

   Now, using identity a³ + b³ = (a + b) (a² – ab + b²)

   [tex]⟹α³ + β³ = (α + β) (α² – αβ + β²)[/tex]

   [tex]⟹α³ + β³ = (\dfrac{-4}{5}) ( \dfrac{-74}{25} – (\dfrac{-9}{5}))[/tex]

   [tex]⟹α³ + β³ = (\dfrac{-4}{5}) ( \dfrac{-74}{25} +\dfrac{9}{5})[/tex]

   [tex]⟹α³ + β³ = (\dfrac{-4}{5}) ( \dfrac{-74+45}{25})[/tex]

   [tex]⟹α³ + β³ = (\dfrac{-4}{5}) ( \dfrac{-29}{25})[/tex]

   [tex]⟹α³ + β³ = \dfrac{-4×(-29)}{5×25}[/tex]

   [tex]{\boxed{⟹α³ + β³ = \dfrac{116}{125}}}[/tex]

   Hence, The value of [tex]{\bold{α³ + β³ \:is\: \dfrac{116}{125}.}}[/tex]

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2. Given equation

   5x² + 4x – 9 = 0

   To find the value of

   α³ + β³

   By comparing with

   ax² – bx + c = 0

   we get

   a = 5 , b = 4 and c = -9

   We know that

   Sum of the zeroes = α + β = -b/a

   Product of the zeroes = αβ = c/a

   we have

   α + β = -4/5

   αβ = -(-9)/5 = 9/5

   now we have to find

   α³ + β³

   We know that

   (α + β)³ = α³ + β³ + 3αβ( α + β)

   (α + β)³ – 3αβ( α + β) = α³ + β³

   Put the value

   (α + β)³ – 3αβ( α + β)

   (-4/5)³ – 3×9/5(-4/5)

   -64/125 – 27/5(-4/5)

   -64/125 + 108/25

   -64/125 + 108×5/125

   -64/125 + 540/125

   476/125

   Answer

   476/125

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