Question 235 of 455HardMCQ
The number of pairs (a,b) of real numbers such that whenever α is a root of x2+ax+b=0, α2−2 is also a root of this equation, is
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Official solution
Consider the equation x2 + ax + b = 0
If has two roots (not necessarily real \[\alpha\] \& \[\beta\])
Either \[\alpha\] = \[\beta\] or \[\alpha\] \[\ne\] \[\beta\]
Case (1) If \[\alpha\] = \[\beta\], then it is repeated root. Given that \[\alpha\]2 - 2 is also a root
So, \[\alpha\] = \[\alpha\]2 - 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 \[\alpha\] \[\ne\] \[\beta\]
Then
(I) \[\alpha\] = \[\alpha\]2 - 2 and \[\beta\] = \[\beta\]2 - 2
Hence, (a, b) = (-(\[\alpha\] + \[\beta\]), \[\alpha\] β)
(−1,−2)
(II)α=β2−2 andβ=α2−2
Thenα−β=β2−α2 = (β−α) (β+α)
Sinceα=βwe getα+β=β2 +α2−4α+β= (α+β)2−2α\[\beta\] - 4
Thus -1 = 1 -2 \[\alpha\] β−4 which impliesα\[\beta\] = -1 Therefore (a, b) = (-(\[\alpha\] + \[\beta\]), \[\alpha\] β)
= (1,−1)
(III)α=α2−2 =β2−2 andα=β⇒α=−βThusα= 2,β=−2α=−1,β= 1
Therefore (a, b) = (0,-\[4) \& (0, 1)
(IV) \]\beta=α2−2 =β2−2 andα=\beta\[ is same as (III) Therefore we get 6 pairs of (a, b)
Which are (2, 1), (\]-4, 4), (−1,−2), (1,−1), (0,−
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