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

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 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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