Find a ^3+ ß^3 , ifa and B are the zeroes of the polynomial
2 x^2 -x + 5​

2 thoughts on “Find a ^3+ ß^3 , ifa and B are the zeroes of the polynomial<br />2 x^2 -x + 5​”

1. Answer:

   Given :-

   • α and β are the zeros of the polynomial 2x² – x + 5.

   To Find :-

   • What is the value of α³ + β³.

   Solution :-

   Given equation :

   [tex] \longmapsto[/tex] [tex] \sf 2{x}^{2} – x + 5 =\: 0[/tex]

   where,

   • a = 2
   • b = – 1
   • c = 5

   Now, we have to find the sum of two roots :

   [tex] \mapsto \sf\boxed{\bold{\pink{\alpha + \beta =\: – \dfrac{b}{a}}}}[/tex]

   Then,

   [tex] \implies \sf \alpha + \beta =\: – \dfrac{- 1}{2}[/tex]

   [tex] \implies \sf\bold{\purple{\alpha + \beta =\: \dfrac{1}{2}}}[/tex]

   Again, we have to find the product of two roots :

   [tex] \mapsto \sf\boxed{\bold{\pink{\alpha\beta =\: \dfrac{c}{a}}}}[/tex]

   Then,

   [tex] \implies \sf\bold{\purple{\alpha\beta =\: \dfrac{5}{2}}}[/tex]

   Now, we have to find the value of α³ + β³ :

   As we know that :

   ➦ a³ + b³ = (a + b)³ – 3ab(a + b)

   According to the question by using the formula we get :

   [tex] \leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: {(\alpha + \beta)}^{3} – 3\alpha\beta(\alpha + \beta)\\[/tex]

   [tex] \leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: {\bigg(\dfrac{1}{2}\bigg)}^{3} – 3\bigg(\dfrac{5}{2}\bigg)\bigg(\dfrac{1}{2}\bigg)\\[/tex]

   [tex] \leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: \dfrac{1}{8} – 3 \times \dfrac{5}{2} \times \dfrac{1}{2}\\[/tex]

   [tex] \leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: \dfrac{1}{8} – 3 \times \dfrac{5}{4}\\[/tex]

   [tex] \leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: \dfrac{1}{8} – \dfrac{15}{4}\\[/tex]

   [tex] \leadsto \sf {\alpha}^{3} + {\beta}^{3} =\: \dfrac{1 – 30}{8}\\[/tex]

   [tex] \leadsto \sf\bold{\red{{\alpha}^{3} + {\beta}^{3} =\: \dfrac{- 29}{8}}}\\[/tex]

   [tex] \sf\boxed{\bold{\green{\therefore\: The\: value\: of\: {\alpha}^{3} + {\beta}^{3}\: is\: \dfrac{- 29}{8}.}}}[/tex]

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2. Answer:-

   Given:-

   α and β are the roots of 2x² – x + 5.

   On comparing the given polynomial with the standard form of a quadratic equation i.e., ax² + bx + c = 0 we get;

   a = 2

   b = – 1

   c = 5.

   We know that;

   Sum of the roots = – b/a

   ⟹ α + β = – ( – 1) / 2

   ⟹ α + β = 1/2 — equation (1)

   And,

   product of the roots = c/a

   ⟹ αβ = 5/2 — equation (2)

   Now,

   we have to find:-

   ⟹ α³ + β³

   We know that;

   a³ + b³ = (a + b)³ – 3ab(a + b)

   Hence;

   ⟹ α³ + β³ = (α + β)³ – 3αβ(α + β)

   Putting the respective values from equations (1) and (2) we get;

   ⟹ α³ + β³ = (1/2)³ – 3(5/2)(1/2)

   ⟹ α³ + β³ = (1/8) – (15/4)

   ⟹ α³ + β³ = (1 – 30)/8

   ⟹ α³ + β³ = – 29/8

   ∴ The value of α³ + β³ is – 29/8.

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