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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Nidhin Kurian
·
4 years, 3 months ago

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@Nidhin Kurian
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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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Calvin Lin
Staff
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4 years, 3 months ago

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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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Samuel Queen
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4 years, 3 months ago

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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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Nidhin Kurian
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4 years, 3 months ago

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its, not a quadratic equation as such because it has 3 real roots
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Nidhin Kurian
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4 years, 3 months ago

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@Nidhin Kurian
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yah! u r right its an identity............ just for confusion.
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Diksha Verma
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4 years, 3 months ago

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I wonder. Did you get this from Praveen Tyagi's FB profile?
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Parth Kohli
·
4 years, 3 months ago

@Riya Gupta
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Because if you take lcm of the denominators and expand, and do some algebra there is no x left, infact u get 1=1
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Aditya Parson
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4 years, 3 months ago

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@Aditya Parson
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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
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Riya Gupta
·
4 years, 3 months ago

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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; – Nidhin Kurian · 4 years, 3 months ago

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– Calvin Lin Staff · 4 years, 3 months ago

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.Log in to reply

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. – Samuel Queen · 4 years, 3 months ago

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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. – Nidhin Kurian · 4 years, 3 months ago

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its, not a quadratic equation as such because it has 3 real roots – Nidhin Kurian · 4 years, 3 months ago

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– Diksha Verma · 4 years, 3 months ago

yah! u r right its an identity............ just for confusion.Log in to reply

I wonder. Did you get this from Praveen Tyagi's FB profile? – Parth Kohli · 4 years, 3 months ago

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– Diksha Verma · 4 years, 3 months ago

??Log in to reply

– Parth Kohli · 4 years, 3 months ago

Look, you just liked my comment. lolLog in to reply

yes – Nidhin Kurian · 4 years, 3 months ago

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– Riya Gupta · 4 years, 3 months ago

can u write the solution plsss....Log in to reply

the roots are -a,-b,-c, assuming a, b, c are distinct real numbers – Nidhin Kurian · 4 years, 3 months ago

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– Riya Gupta · 4 years, 3 months ago

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

– Aditya Parson · 4 years, 3 months ago

You got 1 on both sides?Log in to reply

– Riya Gupta · 4 years, 3 months ago

i am not getting a clear solutionLog in to reply

– Aditya Parson · 4 years, 3 months ago

Because if you take lcm of the denominators and expand, and do some algebra there is no x left, infact u get 1=1Log in to reply

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

this equation is forming an identity i guess – Riya Gupta · 4 years, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

1=1 implies 1-1=0 so 0=0 :P right?Log in to reply

– Riya Gupta · 4 years, 3 months ago

very funny.......ha! :/Log in to reply