If α and β are the zeros of the polynomial f(x)=5x^2+4x−9 then evaluate the following: alpha^3+beta^3 About the author Mary
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] ━━━━━━━━━━━━━━━━━━ Reply
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 Reply
Given:–
To Find:–
Formula used:-
Solution:–
Comparing the given equation with standard equation i.e. ax² + bx + c = 0 we get,
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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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