If a, b, c are distinct real numbers, then, the number of real roots of \frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-a)(x-c)}{(b-a)(b-c)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}-1 is?
62to be more lucid.. what does this show? (The above equation is not a quadratic equation?)
Actually you got into the trap of the question.. and after getting that answer that you did, something must strike you!
1Most true . Then it's not an equation but an Identity. Right ?
62Bingo :)
Lets hope that the others take a cue from this trick.. and I will post the next question.. (open for everyone but aveek :) :D
62No one else!! I thought this was really important for beginners to understand polynomials and quadratics...
Neone other than Aveek, please give the complete proof.. (just read teh posts 2,3,4,5 )
11The roots are 3 and they are a,b,c
See
According to langeranges interpolation
F(x) = (x-a)(x-b)F(c)/(c-a)(c-b) + (x-c)(x-b)F(a)/(a-c)(a-b) + (x-a)(x-c)F(b)/(b-a)(b-c)
Now in the above equation F(c) = F(a) = F(b) = 1 and hence the root of the above equation will be a,b,c
24I odnt get what are we discussing here..............
putting x=a,x=b,x=c gives the expression zero..so its not an eqn..its an identity.......
and can have any number of roots..
62So what is it? and y do you only think?
there have been 3-4 questions of the same kind put here.. may be u were all not reading them or understanding them :(
23its an identity ..as simple as dat ....
its like a2 -b 2 = (a+b)(a-b) is an identity ...same is dis ....
39already this one had a healthy discussion here, not able to give the old link as of now...
anwyays nishant sir .. everyr eal number is a root of this equation thats is what it means.
