# FINDING QUADRATIC EQUATION WITH THE ROOTS IN TERMS OF ALPHA BETA

Finding Quadratic Equation with the Roots in Terms of Alpha Beta :

Here we are going to see some example problems of finding quadratic equation with the roots in terms of alpha beta.

## Formulas of Alpha Beta in Quadratic Equation

(α2 + β2)  =  (α + β)- 2αβ

(α3 - β3)  =  (α - β)+ 3αβ(α - β)

(α4 + β4)  =  (α2β2)- 2α2β2

α - β  =  √[(α + β)- 4αβ]

## Finding Quadratic Equation with the Roots in Terms of Alpha Beta - Questions

Question 1 :

The roots of the equation x2 + 6x − 4 = 0 are α, β. Find the quadratic equation whose roots are

(i) 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 βas roots

x2 - (α2 + β2)x + α2 β2 = 0

x2 - (α2 + β2)x + (αβ)2 = 0  ---(1)

α2 + β=  (α + β)2 - 2αβ  =  (-6)- 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

(ii)  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

(iii) 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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