Look Closely...it's a quadratic equation. Now find the number of solutions it has.
4 years, 5 months ago
take L.H.S=f(x), and it is easy to show that f'(x)=0; for all x. ie, f(x)= k = f(-a)= 1, so for all x f(x)=1;
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That is a good observation. A simpler approach would be to make common denominator and consider the quadratic polynomial in \(x\), and easily see that the quadratic coefficient and linear coefficient are both 0.
If the L.H.S=f(x) (if none of the a,b,c's are equal to each other) then f(x)=1 for all x in R. So f(x)-1=0 for all x in R. Thus f(x)-1=p(x) is the zero polynomial and not a quadratic.
yes, my answer of -a, -b, -c is not correct as indeed it is an identity function the equation is satisfied for all x belonging to R.
its, not a quadratic equation as such because it has 3 real roots
yah! u r right its an identity............ just for confusion.
I wonder. Did you get this from Praveen Tyagi's FB profile?
Look, you just liked my comment. lol
can u write the solution plsss....
the roots are -a,-b,-c, assuming a, b, c are distinct real numbers
how did you got the answer? plzzz explain....i'm not able to solve! :|
You got 1 on both sides?
i am not getting a clear solution
Because if you take lcm of the denominators and expand, and do some algebra there is no x left, infact u get 1=1
arey.....i took the denominator on the r.h.s. and got 0=0
but m wondering how to get solutions......
this equation is forming an identity i guess
1=1 implies 1-1=0 so 0=0 :P right?
very funny.......ha! :/