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.

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TopNewesttake 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.

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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.

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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.

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its, not a quadratic equation as such because it has 3 real roots

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yah! u r right its an identity............ just for confusion.

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I wonder. Did you get this from Praveen Tyagi's FB profile?

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??

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Look, you just liked my comment. lol

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yes

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can u write the solution plsss....

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the roots are -a,-b,-c, assuming a, b, c are distinct real numbers

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how did you got the answer? plzzz explain....i'm not able to solve! :|

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You got 1 on both sides?

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but m wondering how to get solutions......

this equation is forming an identity i guess

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