Here we are going to see some example problems of finding quadratic equation with the roots in terms of alpha beta.
(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 1 :
The roots of the equation x2 + 6x − 4 = 0 are α, β. Find the quadratic equation whose roots are α2 and β2.
Solution :
x2 + 6x − 4 = 0
a = 1, b = 6 and c = -4
Sum of roots (α + β) = -b/a = -6/1 = -6
Product of roots (α β) = c/a = -4/1 = -4
α = α2, β = β2
General form of quadratic equation with α and β as roots
x2 - (α + β)x + αβ = 0
Required quadratic equation with α2 and β2 as roots
x2 - (α2 + β2)x + α2 β2 = 0
x2 - (α2 + β2)x + (αβ)2 = 0 ---(1)
α2 + β2 = (α + β)2 - 2αβ = (-6)2 - 2(-4)
α2 + β2 = 36 + 8 = 44
By applying the value of α2 + β2 in (1)
x2 - 44x + (-4)2 = 0
x2 - 44x + 16 = 0
Problem 2 :
The roots of the equation x2 + 6x − 4 = 0 are α, β. Find the quadratic equation whose roots are 2/α and 2/β.
Solution :
Required quadratic equation with roots 2/α and 2/β
x2 - [(2/α) + (2/β)]x + (2/α) (2/β) = 0
x2 - [(2β + 2α)/αβ] x + (4/αβ) = 0
x2 - 2[(α+β)/αβ] x + (4/αβ) = 0
x2 - 2[(-6)/(-4)] x + (4/(-4)) = 0
x2 - 3x - 1 = 0
Problem 3 :
The roots of the equation x2 + 6x − 4 = 0 are α, β. Find the quadratic equation whose roots are α2β and β2α.
Solution :
Required quadratic equation with roots 2/α and 2/β
x2 - [(α2β) + (β2α)]x + (α2β) (β2α) = 0
x2 - αβ (α + β) x + α3β3 = 0
x2 - αβ (α + β) x + (αβ)3 = 0
x2 - (-4) (-6) x + (-4)3 = 0
x2 - 24x - 64 = 0
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