# 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]

## Solved Questions

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 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

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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