39wats the proof u have got?
The equation ax3 +bx2 +cx+d = 0 is known to have three distinct
real roots.
How many real roots are there of the equation
4(ax3 + bx2+ cx + d)(3ax + b) = (3ax2 + 2bx + c)2?
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11 Answers
39sir i felt so. if it comes very much in syllabus, then no probs. ( and then sorry for the title)
1the equation becomes
3a2x4+6cax2+12dax+4bax3+4db-c3=0
now let this b f(x) in wich der is only 1 sign change so it has only 1 +ve real root........now put f(-x)=der will b 3 sign change so 3 -ve roots.........sorry i posted 2 -ve real roots..........edited
9this is jee syllabus only b555
take y =ax3 +bx2+cx+d
now take g(x) = y'2-2y"y'
we want g(x) =0
g'(x)=2y'''y =12ay
now minima and maxima of g(x) are at g'(x)=0 ie y=0
moreover at extremum g(x) = y'2 (as y=0) which is >0
also note that g(x) = 3a2x4 + ......... which is positive as x→ ±∞
all these imply curve g(x) is always above x axis ie no real solution for g(x)=0
341What I had in mind was:
Let y = ax3+bx2+cx+d. Then the equation can be written as:
2yy" = (y')2
Now y has three distinct roots implies that it does not share any roots with y'. Also it implies that y' has two distinct roots. The possibility we that y' and y" share a root is hence ruled out as that would mean that y' has only one root.
Thus y' has no roots in common with y or y". But the given equation implies that whenever y' = 0 either y = 0 or y" = 0, which is a contradiction and we are done
9but prophet sir u ve proved that
2yy" ≠(y')2 only when y or y'' or y" is 0 what abt other cases ???