If α and β are the roots of a quadratic polynomial 3x²-6x-1, find the values of
I) α and β
2) α²+β²
3) α-β
4) α³+β³
5) α³-β³
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P(x) = 3x^2-6x-1 = ax^2+bx+c=0
α+β=-b/a=-(-6)/3=2
α+β=2 --------------eqn(1)
now,
αβ=c/a=-1/3 -------------eqn(2)
we know that, (α+β)^2-(α-β)^2=4*αβ
(2)^2-(α-β)^2=4*(-1/3)
4-(α-β)^2=-4/3
(α-β)^2=4+4/3
(α-β)^2=16/3
α-β = √(16/3)
α-β = +-4/√3 ----------------eqn(3)
eqn(1) + eqn(3)
2α = 2+-4/√3
α = 1+-2/√3
from eqn (1)
β=2-α
β=2-(1+-2/√3)
β=1+2/√3
when we put the value of α and β ,then we can find the value of
α^2+β^2 , α^3+β^3, and α^3-β^3
α+β=-b/a=-(-6)/3=2
α+β=2 --------------eqn(1)
now,
αβ=c/a=-1/3 -------------eqn(2)
we know that, (α+β)^2-(α-β)^2=4*αβ
(2)^2-(α-β)^2=4*(-1/3)
4-(α-β)^2=-4/3
(α-β)^2=4+4/3
(α-β)^2=16/3
α-β = √(16/3)
α-β = +-4/√3 ----------------eqn(3)
eqn(1) + eqn(3)
2α = 2+-4/√3
α = 1+-2/√3
from eqn (1)
β=2-α
β=2-(1+-2/√3)
β=1+2/√3
when we put the value of α and β ,then we can find the value of
α^2+β^2 , α^3+β^3, and α^3-β^3
chrisle:
Thank you so much Karthik4297!
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