1arey chhoro naa woh sab...
these methods are very helpful at times... u cant deny akshay...
but,,, ans 'b" hai naa?
Find all the values of m for which equation
(x2+x+1)2-(m-3)(x2+x+1)+m=0 (where m is a rel parameter) will have real roots
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38 Answers
11i'm not finding it useful as well, vishal. So forget about it[4]
1hehehe akshay
bt b a game bro i m against solving like this too bt if IIT likes us to solve it like this by making the paper multiple choice then no1 shud avoid using these methods
111Will it be helpful if we take x2+x+1 as (x+1/2)2+3/4, then put x+1/2 as X and then proceed?
1yes Answer is (b)
I myself use these methods but in the xams or my Test Series,not when I have time
1if ur finding it helpful then it might b useful i will just cheak it though
1i dun think so..
becozzzz......... have you thot of your next step ?
wat will u do after taking that thing that u hav taken....
12 ram dude i din assume nethng just put m=0 i m not a person who like to xactly solve the Q
:D
1prophet was with d so this is for him
put m=-17 & the condition wll b satisfied hence ur ans is wroooonggg :D
1sry prophet the post get deleted i will give it again tommorow itna lamba lamba phir likhane ka man nahi kar raha
62(x2+x+1)2-(m-3)(x2+x+1)+m=0
x2+x+1 to have a real root should be
x2+x+1/4 + 3/4
should thus have a vlue greater than equal to 3/4
so
y2-(m-3)y+m = 0 should have one root greater than equal to 3/4
I guess this should suffice?
1x^2 + x+ 1 =y
two positive real roots by sridharcharya..
tell this c1, c2 (only ± will be there so let it be one variable 'c')
now x^2 + x+ 1 =c
thsi should have real roots..
so D>=0 for this..
1u just did D>0 for the equation
Don u think I would'nt have given this question,if it was so easy
1for the equation f(t) where t=x2+x+1
right??
think more then
I will not tell the answer or u will think the reverse way
1Here are the options
a)(-infinity,1)U(9,infinity)
b)(-infinity,-45/4)U(9,infinity)
c)(-infinity,1)
d)(9,infinity)