SUM AND PRODUCT OF ROOTS OF QUADRATIC EQUATION PROBLEMS

(α2 + β2)  =  (α + β)- 2αβ

(α3 - β3)  =  (α - β)+ 3αβ(α - β)

(α4 + β4)  =  (α2β2)- 2α2β2

α - β  =  √[(α + β)- 4αβ]

Problem 1 :

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)- 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)  =  (α2β2)/αβ

=  [(α + β)2 - 2αβ]/αβ

=  [(7/2)- 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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