(α2 + β2) = (α + β)2 - 2αβ
(α3 - β3) = (α - β)3 + 3αβ(α - β)
(α4 + β4) = (α2 + β2)2 - 2α2β2
α - β = √[(α + β)2 - 4αβ]
Problem :
Write each of the following expression in terms of α + β and α β
(i) (α / 3β) + (β / 3α)
Solution :
= (α / 3β) + (β / 3α)
= (3α2 + 3β2) / 9αβ
= 3(α2 + β2) / 9αβ
= 3[(α + β)2 - 2αβ] / 9αβ
= [(α + β)2 - 2αβ] / 3αβ
(ii) (1 / α2β) + (1 / β2α)
Solution :
(1 / α2β) + (1 / β2α) = (β + α) / α2β2
= (α + β) / (αβ)2
(iii) (3α - 1) (3β - 1)
Solution :
(3α - 1) (3β - 1) = 9αβ - 3α - 3β + 1
= 9αβ - 3(α + β) + 1
(iv) [(α + 3)/β] + [(β + 3)/α]
Solution :
[(α + 3)/β] + [(β + 3)/α] = [α(α + 3) + β(β + 3)] / αβ
= (α2 + 3α + β2 + 3β)/αβ
= (α2 + β2 + 3(α + β))/αβ
= [(α + β)2 - 2 α β + 3(α + β)]/αβ
Problem 2 :
The roots of the equation 2x2 −7x + 5 = 0 are α and β. Without solving for the roots, find
(i) (1/α) + (1/β)
Solution :
a = 2, b = -7 and c = 5
Sum of roots (α + β) = -b/a = -(-7)/2 = 7/2
Product of roots (α β) = c/a = 5/2
(1/α) + (1/β) = (β + α)/αβ
= (7/2)/(5/2)
(1/α) + (1/β) = 7/5
(ii) (α/β) + (β/α)
Solution :
Sum of roots (α + β) = -b/a = -(-7)/2 = 7/2
Product of roots (α β) = c/a = 5/2
(α/β) + (β/α) = (α2 + β2)/αβ
= [(α + β)2 - 2αβ]/αβ
= [(7/2)2 - 2(5/2)]/(5/2)
= [(49/4) - 5]/(5/2)
= [(49 - 20)/4]/(5/2)
= (29/4)/(5/2)
= 29/10
(iii) [(α + 2)/(β + 2)] + [(β + 2)/(α + 2)]
Solution :
[(α + 2)/(β + 2)] + [(β + 2)/(α + 2)]
= [(α + 2)2 + (β + 2)2]/(α + 2)(β + 2)
= (α2 + 4α + 4 + β2 + 4β + 4)/(α β + 2α + 2β + 4)
= (α2 + β2 + 4(α + β) + 8)/(α β + 2(α + β) + 4)
= (29/4) + 4(7/2) + 8)/((5/2) + 2(7/2) + 4)
= (29/4) + 14 + 8)/((5 + 14 + 8)/2)
= ((29 + 88)/4)/((5 + 14 + 8)/2)
= (117/4)/(27/2)
= 13/6
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