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Question

If α and β are the zeros of the polynomial 2x24x+5, find the value of
(i) α2+β2
(ii) 1α+1β
(iii) (αβ)2
(iv) α3+β3

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Solution

2x24x+5
α+β=ba=(4)2=42=2
αβ=ca=52
(α2+β2)=(α+β)22αβ
=45=1
1α+1β=α+βαβ=252=45
(αβ)2=(α2+β2)2αβ
=12(52)=15
=6
(α3+β3)=(α+β)(α2+β2αβ)
=(2)(152)
=2(72)=7

1215228_1354167_ans_46af00fe5e964c3eabd52ffb88191046.jpg

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