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If α & ß are the zeroes of the polynomial 2x2 - 4x + 5, then find the value of a.α2 + ß2 b. 1/ α + 1/ ß c. (α - ß)2 d. 1/α2 + 1/ß2 e. α3 + ß3 (Ans:-1, 4/5 ,-6, - 4/25 ,-7)
Ans: Let p(x) = 2x2 - 4x +5
α+β = - b/a = 4/2 = 2
αβ = c/a = 5/2
a) α2+β2 = (α+β)2 - 2αβ
Substitute to get = α2+β2 = -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) α3+β3 = (α+β)(α2+β2 - αβ)
Substitute this,
to get, α3+β3 = -7
Factorize x squared + 6x + 8
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