If alpha nd beta are the zeroes of polynomial 2x^2-7x-3 then find the value of alpha^3+beta^3

1 thought on “If alpha nd beta are the zeroes of polynomial 2x^2-7x-3 then find the value of alpha^3+beta^3”

1. Step-by-step explanation:

   Given:–

   Quadratic pilynomial 2x^2-7x-3

   To find:–

   If alpha nd beta are the zeroes of polynomial 2x^2-7x-3 then find the value of alpha^3+beta^3

   Solution:–

   Given quadratic polynomial = 2x^2-7x-3

   On comparing with the standard quadratic pilynomial ax^2+bx+c then

   a = 2

   b=-7

   c=-3

   If α and β are the zeroes of the pilynomial then

   Sum of the zeroes = -b/a

   =>α+ β = -(-7)/2

   α+ β = 7/2 —————(1)

   Product of the zeroes = c/a

   αβ = -3/2 —————(2)

   On squaring the equation (1) both sides then

   (α+ β)^2 = (7/2)^2

   We know that (a+b)^2 = a^2+2ab+b^2

   =>α^2+2 αβ +β^2 = 49/4

   =>α^2+2(-3/2)+β^2 = 49/4

   =>α^2-3+β^2 = 49/4

   =>α^2+β^2 = (49/4)+3

   =>α^2+β^2 = (49+12)/4

   =>α^2+β^2 = 61/4———–(3)

   Now the value of α^3+ β^3

   We know that

   a^3+b^3 =(a+b)(a^2-ab+b^2)

   α^3+ β^3 = (α+ β)(α^2+β^2-αβ)

   =>α^3+ β^3 = (7/2)[(61/4)-(-3/2)]

   =>α^3+ β^3 = (7/2)[(61/4)+(3/2)]

   =>α^3+ β^3 = (7/2)[(61+6)/4]

   =>α^3+ β^3 = (7/2)(67/4)

   =>α^3+ β^3 = (7×67)/(2×4)

   =>α^3+ β^3 = 469/8

   Answer:–

   The value of α^3+ β^3 for the given problem is

   469/8

   Used formulae:–

   • Sum of the zeroes = -b/a

   • Product of the zeroes = c/a

   • (a+b)^2 = a^2+2ab+b^2

   • a^3+b^3 =(a+b)(a^2-ab+b^2)

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