4@xyz : (a+c)(ax2+2bx+c) = 2(ac-b2)(x2+1)
ax2 + 2bx + c = 0
=>2(x2 + 1) (ac - b2) = 0
If the roots of ax2 + 2bx + c = 0 be possible and different then the roots of
(a+c)(ax2+2bx+c) = 2(ac-b2)(x2+1)
will be impossible and vice-versa.............prove it.
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7 Answers
42(x2 + 1) (ac - b2) = 0
This is impossible since x2 ≠-1 since x ε R & ac ≠b2 as given eqn has distinct roots
Vice Versa can be prooved similarly
(I Hope this is the 2 liner !!!)
41@abv
read this carefully
If the roots of ax2 + 2bx + c = 0 be possible and different then the roots of ...
4@xyz : The qs is (a+c)(ax2+2bx+c) = 2(ac-b2)(x2+1)
will be impossible for the roots at which ax2 + 2bx + c = 0 be possible
That's how I understood it as
if it's wrong then may be i can't think of any other solution right now
21well simple meaning of q is if Discrim of first eq is >=0 then u hav to prove that Discri of second eq is <=0.......thats it!
