# solve the following problem?

if α and β are zeroes of the polynomial 6x^2 + x + 1 , then find the value of :-

(i) α^3 β + α β^3

(ii) α^2 + β^2

(iii) α^2/β + β^2/α

### 2 Answers

Relevance

- Elizabeth MLv 77 years ago
(α+ß)=-b/a=-1/6, and αß=c/a=1/6

(i) α^3 β + α β^3

=αß(α²+ß²)

=αß[(α+ß)²-2αß]

=(1/6)[1/36 - 1/3]

=-11/216

(ii)α²+ß²

=(α+ß)²-2αß

=1/36-1/3

=-11/36

(iii) α²/ß+ß²/α

=(α³+ß³)/(αß)

=(α+ß)((α+ß)²-3αß)/(αß)

=17/36

- moeLv 77 years ago
6x^2 + x + 1 = 0

Factorize to get:

(6x+1)(x+1) = 0

so,

if(6x+1) = 0 then x = -1/6 or

else (x+1) = 0 then x = -1

so, roots are: -1 (=α) and -1/6(=β)

in which case:

(i) α^3 β + α β^3 =(-1)^3(-1/6) + (-1)(-1/6)^3 = 0.1723

(ii) α^2 + β^2 = (-1)^2 + (1/6)^2 = 1.0278

(iii) α^2/β + β^2/α = (-1)^2/(-1/6) + (-1/6)^2/(-1) = -6.0278

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