Question

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

Answer 1

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

Answer 2

Given:-

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

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 = $\dfrac{-b}{a}$

$⟹α + β = \dfrac{-4}{5}$

Product of zeroes = $\dfrac{c}{a}$

$⟹αβ = \dfrac{-9}{5}$

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

$⟹(α + β)² = α² + β² + 2αβ$

Putting values,

$⟹(\dfrac{-4}{5})² = α² + β² + 2×\dfrac{-9}{5}$

$⟹\dfrac{16}{25}= α² + β² - \dfrac{18}{5}$

$⟹α² + β² = \dfrac{16}{25}- \dfrac{18}{5}$

$⟹α² + β² = \dfrac{16-90}{25}$

${\boxed{⟹α² + β² = \dfrac{-74}{25}}}$

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

$⟹α³ + β³ = (α + β) (α² - αβ + β²)$

$⟹α³ + β³ = (\dfrac{-4}{5}) ( \dfrac{-74}{25} - (\dfrac{-9}{5}))$

$⟹α³ + β³ = (\dfrac{-4}{5}) ( \dfrac{-74}{25} +\dfrac{9}{5})$

$⟹α³ + β³ = (\dfrac{-4}{5}) ( \dfrac{-74+45}{25})$

$⟹α³ + β³ = (\dfrac{-4}{5}) ( \dfrac{-29}{25})$

$⟹α³ + β³ = \dfrac{-4×(-29)}{5×25}$

${\boxed{⟹α³ + β³ = \dfrac{116}{125}}}$

Hence, The value of ${\bold{α³ + β³ \:is\: \dfrac{116}{125}.}}$

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