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Algebra of Limits

If α, β are t...

Question

If $α, β$ are the roots of the equation $x^{2} - a x + b = 0$ and roots of the eqaution $b x^{2} - 4 x + 4 = 0$ are $α + \frac{β^{2}}{α}$ , $β + \frac{α^{2}}{β}$ then $a + b$ can be -

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Solution

The correct options are
B $1$
D $3$
$x^{2} + a x + b = 0$
$α β = b, α + β = a$
Roots of the equation $b x^{2} - 4 x + 4 = 0$ are $α + \frac{β^{2}}{α}$ , $β + \frac{α^{2}}{β}$
$α + \frac{β^{2}}{α} = \frac{(α + β)^{2} - 2 α β}{α} = \frac{a^{2} - 2 b}{α} β + \frac{α^{2}}{β} = \frac{a^{2} - 2 b}{β}$
So, the equation having these two roots will be
$x^{2} + (sum of roots) x + (product of roots) = 0$
$\Rightarrow x^{2} - \frac{a (a^{2} - 2 b)}{b} x + \frac{(a^{2} - 2 b)^{2}}{b} = 0 b x^{2} - a (a^{2} - 2 b) x + (a^{2} - 2 b)^{2} = 0............ (i) b x^{2} - 4 x + 4 = 0................ (i i)$
Comparing $(i)$ and $(i i)$ we have
$a^{2} - 2 b = \pm 2 & a (a^{2} - 2 b) = 4 a = 2 & b = 1 a = - 2 & b = 3$

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