Question

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

Answer 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)