The numbers of pairs (a, b) of real numbers, such that whenever \alpha is a root of the equation x2 + ax + b = 0, \alpha2 - 2 is also a root of this equation, is :

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Answer: A. 6 Consider the equation x2 + ax + b = 0 If has two roots (not necessarily real \alpha & \beta) Either \alpha = \beta or \alpha 
e \beta Case (1) If \alpha = \beta, then it is repeated root. Given that \alpha2 - 2 is also a root So, \alpha = \alpha2 - 2 \Rightarrow (\alpha + 1)(\alpha - 2) = 0 \Rightarrow \alpha = -1 or \alpha = 2 When \alpha = -1 then (a, b) = (2, 1) \alpha = 2 then (a, b) = (-4, 4) Case (2) If \alpha 
e \beta Then (I) \alpha = \alpha2 - 2 and \beta = \beta2 - 2 Hence, (a, b) = (-(\alpha + \beta), \alpha\beta) (-1, -2) (II) \alpha = \beta2 - 2 and \beta = \alpha2 - 2 Then \alpha - \beta = \beta2 - \alpha2 = (\beta - \alpha) (\beta + \alpha) Since \alpha 
e \beta we get \alpha + \beta = \beta2 + \alpha2 - 4 \alpha + \beta = (\alpha + \beta)2 - 2\alpha\beta - 4 Thus -1 = 1 -2 \alpha\beta - 4 which implies \alpha\beta = -1 Therefore (a, b) = (-(\alpha + \beta), \alpha\beta) = (1, -1) (III) \alpha = \alpha2 - 2 = \beta2 - 2 and \alpha 
e \beta \Rightarrow \alpha = - \beta Thus \alpha = 2, \beta = -2 \alpha = -1, \beta = 1 Therefore (a, b) = (0, -4) & (0, 1) (IV) \beta = \alpha2 - 2 = \beta2 - 2 and \alpha 
e \beta is same as (III) Therefore we get 6 pairs of (a, b) Which are (2, 1), (-4, 4), (-1, -2), (1, -1), (0, -4) Option (a)

The numbers of pairs (a, b) of real numbers, such that whenever $$\alpha$$ is a root of the equation x2 + ax + b = 0, $$\alpha$$2 $$-$$ 2 is also a root of this equation, is :

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