(α2 + β2) = (α + β)2 - 2αβ
(α3 - β3) = (α - β)3 + 3αβ(α - β)
(α4 + β4) = (α2 + β2)2 - 2α2β2
α - β = √[(α + β)2 - 4αβ]
Question 1 :
If α, β are the roots of 7x2+ax+2=0 and if β − α = −13/7 Find the values of a.
Solution :
7x2 + ax + 2=0
a = 7, b = a and c = 2
Sum of roots (α + β) = -b/a = -a/7
Product of roots (α β) = c/a = 2/7
β − α = −13/7
-(α - β) = -13/7
(α - β) = -13/7
√(α + β)2 - 4αβ = -13/7
(α + β)2 - 4αβ = (-13/7)2
(-a/7)2 - 4(2/7) = 169/49
(a2/49) - (8/7) = 169/49
a2 - 56 = 169
a2 = 225
a = ±15
Hence the values of a are -15 and 15.
Question 2 :
If one root of the equation 2y2 − ay + 64 = 0 is twice the other then find the values of a.
Solution :
Let α and β are two roots.
α = 2β, β = β
Sum of roots (α + β) = -b/a = -(-a)/2 = a/2
Product of roots (α β) = c/a = 64/2 = 32
α + β = a/2 2β + β = a/2 3β = a/2 β = a/6 ---(1) |
2β(β) = 32 2β2 = 32 β2 = 16 β = ±4 |
When β = 4
4 = a/6
a = 24
when β = -4
-4 = a/6
a = -24
Question 3 :
If one root of the equation 3x2 + kx + 81 = 0 (having real roots) is the square of the other then find k.
Solution :
α = β2
Sum of roots (α + β) = -b/a = -k/3
Product of roots (α β) = c/a = 81/3 = 27
α β = 27 β2β = 27 β3 = 33 β = 3 |
α + β = -k/3 --(1) β2+ β = -k/3 32+ 3 = -k/3 12 = -k/3 -k = 36 k = -36 |
Hence the value of k is -36.
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