If α & ß are the zeroes of the polynomial 2x^{2} - 4x + 5, then find the value of a.α^{2} + ß^{2} b. 1/ α + 1/ ß c. (α - ß)^{2} d. 1/α^{2} + 1/ß^{2} e. α^{3} + ß^{3 }(Ans:-1, 4/5 ,-6, - 4/25 ,-7)
Ans: Let p(x) = 2x^{2} - 4x +5
α+β = - b/a = 4/2 = 2
αβ = c/a = 5/2
a) α^{2}+β^{2} = (α+β)^{2} - 2αβ
Substitute to get = α^{2}+β^{2} = -1
b) 1/a + 1/β = α + β/αβ
substitute , then we get = 1/a + 1/β = 4/5
b) (α-β)^{2} = (α+β)^{2} - 4 αβ
Therefore we get, (α-β)^{2} = - 6
d) 1/α ^{2} + 1/β ^{2 }= α ^{2} + β ^{2}/ αβ ^{2} = - 1/(5/2)^{2}
∴ 1/α ^{2} + 1/ β ^{2} = - 4/25
e) α^{3}+β^{3} = (α+β)(α^{2}+β^{2} - αβ)
Substitute this,
to get, α^{3}+β^{3} = -7