α2+β2 = (α+β)2 - 2αβ
α-β = √(α+β)2 - 4αβ
α3-β3 = (α-β)3 + 3αβ(α-β)
α4+β4 = (α2+β2)2 - 2α2β2
Question 1 :
If α and β are the roots of the equation
3x2-5x+2 = 0
then find the values of
(i) (α/β) + (β/α)
(ii) α - β
(iii) (α²/β) + (β²/α)
Solution :
By comparing the given quadratic equation with the general form of quadratic equation, we get
a = 3, b = -5 and c = 2
Sum of roots : α+β = -b/a α+β = -(-5)/3 α+β = 5/3 |
Product of roots : αβ = c/a αβ = 2/3 |
(i) (α/β)+(β/α)
By combining the above fractions, we get
(α/β)+(β/α) = (α2+β2)/αβ -----(1)
α2+β2 = (α+β)2-2αβ
= (5/3)2-2(2/3)
= (25/9)-(4/3)
= (25-12)/9
α2+β2 = 13/9
By applying the values in (1), we get
(α/β) + (β/α) = (α2+β2)/αβ
= (13/9)/(2/3)
(α/β) + (β/α) = 13/6
(ii) α - β
α-β = √(α+β)2 - 4αβ
= √(5/3)2-4(2/3)
= √(25/9)-(8/3)
= √1/9
α-β = ± 1/3
(iii) (α2/β) + (β2/α)
By combining the given fractions, we get
(α2/β) + (β2/α) = (α3+β3)/αβ ----(1)
α3-β3 = (α-β)3 + 3αβ(α-β)
= (5/3)3-3(2/3)(5/3)
= (125/27)-(10/9)
α3-β3 = 95/27
By applying the values in (1), we get
(α2/β) + (β2/α) = (95/27)/(2/3)
(α2/β) + (β2/α) = 95/18
Question 2 :
If α and β are the roots of
3x2-6x+4 = 0
find the value of α2+β2
Solution :
a = 3 b = - 6 and c = 4
α2+β2 = (α+β)2 - 2αβ ----(1)
Sum of roots : α+β = -b/a = -(-6)/3 α+β = 2 |
Product of roots : αβ = c/a αβ = 4/3 |
By applying the values in (1), we get
= 22-2(4/3)
= 4-(8/3)
= 4/3
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