Question

find alphacube + beta cube, if alpha and beta are the zeroes of the polyinomial 2x^2-x+5

Answer 1

Solution -

Given polynomial,

• 2x² - x + 5

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It is given that α and β are the zeroes of the given polynomial. Firstly, we have to find the α + β.

• Sum of zeroes = $\sf{\dfrac{-b}{a}}$

$\tt\dashrightarrow{\alpha + \beta = \dfrac{-(-1)}{2}}$

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$\tt\dashrightarrow{\alpha + \beta = \dfrac{1}{2}}$

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Similarly,

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

$\tt\dashrightarrow{\alpha \beta = \dfrac{5}{2}}$

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Now,

• α² + β² = (α + β)² - 2αβ

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$\tt\dashrightarrow{\alpha^2 + \beta^2 = \bigg( \dfrac{1}{2} \bigg)^2 - 2 \times \dfrac{5}{2}}$

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$\tt\dashrightarrow{\alpha^2 + \beta^2 = \dfrac{1}{4} - 5}$

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$\tt\dashrightarrow{\alpha^2 + \beta^2 = \dfrac{1 - 20}{4}}$

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$\tt\dashrightarrow{\alpha^2 + \beta^2 = \dfrac{-19}{4}}$

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Now, we need to find

• α³ + β³

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$\bf{\longmapsto{\boxed{\pink{\alpha^3 + \beta^3 = ( \alpha + \beta ) (\alpha^2 + \beta^2 - \alpha \beta )}}}}$

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$\tt\longrightarrow{\alpha^3 + \beta^3 = \bigg( \dfrac{1}{2} \bigg) \bigg( \dfrac{5}{2} - \dfrac{(-19)}{4} \bigg)}$

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$\tt\longrightarrow{\alpha^3 + \beta^3 = \bigg( \dfrac{1}{2} \bigg) \bigg( \dfrac{5}{2} + \dfrac{19}{4} \bigg)}$

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$\tt\longrightarrow{\alpha^3 + \beta^3 = \bigg( \dfrac{1}{2} \bigg) \bigg( \dfrac{10 + 19}{4} \bigg)}$

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$\tt\longrightarrow{\alpha^3 + \beta^3 = \dfrac{1}{2} \times \dfrac{29}{4}}$

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$\bf\longrightarrow{\purple{\alpha^3 + \beta^3 = \dfrac{29}{8}}}$

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$\: \: \: \underline{\sf{Thus,\: value \: of\: \alpha^3 + \beta^3\: is \: \dfrac{29}{8}}}$