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Question

α and β are the roots of ax2+bx+c=0 and ax2+bx+c=0, then prove that f(α).f(β)=1a2[(c1aca1)2(ab1a1b)(bc1b1c)]

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Solution

f(α)=a1α2+b1α+c1
or a1α2+b1α+[c1f(α)]=0.......(1)
Again α is a root of ax2+bx+c=0
aα2+bx+c=0........(2)
solving (1) and (2) by cross multiplication, we get
α2b1cbc1+bf(α)=αac1ca1af(α)=1a1bab1
[ac1a1caf(α)]2=(a1bab1)[b1cbc1+bf(α)]
Writing above as a quadratic in f(α)
a2[f(α)]2+f(α)[]+(ac1a1c)2(a1bab1)(b1cbc1)=0
Above shows that f(α) is a root of
a2t2+t[]+(ac1a1c)2(a1bab1)(b1cbc1)=0...........(1)
Similarly was can assume that f(β) will also be a root of (1). Hence the roots of t quadratic given by (1) are f(α) and f(β).
f(α).f(β)= Product of roots of (1).
or f(α).f(β)=1a2[(ac1a1c)2(a1bab1)(b1cbc1)]........(2)
It is same as given result if we change the signs of both the last brackets in the numerator.
In case α or β be a common root of
ax2+bx+c=0 and f(x)=a1x2+b1x+c=0
then f(α)=0.
Hence from [2], $
f(\alpha ).f(\beta )=0$
(ac1a1c)2(a1bab1)(b1cbc1)=0
or (ac1a1c)2=(bc1b1c)(ab1a1b)
is the required condition.

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