Show that the equation
$$\frac{x^5}{120}+\frac{x^4}{24}+\frac{x^3}{6}+\frac{x^2}{2}+x+1=0$$has no repeated roots.

Question

Show that the equation 
$$\frac{x^5}{120}+\frac{x^4}{24}+\frac{x^3}{6}+\frac{x^2}{2}+x+1=0$$has no repeated roots.

shubhamsrg · Answer

the ques is quite similar to Q.9 of http://www.cmi.ac.in/admissions/sample-qp/ugmath2010.pdf i have the solution and that might sound interesting,,i just solved this paper yesterday! :D

kainui · Answer

Looks like it has something to do with its derivatives.

shubhamsrg · Answer

if f(x) = x^5/5! + x^4/4! ...... x +1 
now x=0 is not a solution for f(x)=0..
then f'(x) = x^4/4! + x^3/3! .... x + 1
i.e. f'(x) = f(x) - x^5/5! .................(1)

now let f(x) = ( (x-a)^m ) * g(x)
where m is some natural no. > 1
it means a is a repeated root of f(x)
now f'(x)= ( m(x-a)^(m-1) ) *g(x) + g'(x) *  ( (x-a)^m ) = (x-a) * (something)
hence a is also a root of f'(x)

coming back to eqn (1),
for x= a , we have

f'(a) = f(a) - a^5/5!
=>0 =  0 - a^5 /5!
=>a=0

which states a=0 is the repeated root but its a contradiction as x=0 is not a solution for f(x) =0

hence no such a exists or in other words,,f(x) has no repeated roots!!

shubhamsrg · Answer

for soln of q.9 of that pdf file,,you just generalize ..

anonymous · Answer

http://finedrafts.com/files/Larson%20PreCal%208th/Larson%20Precal%20CH2.pdf pp. 166

anonymous · Answer

nice work... @shubhamsrg 

let  $x=a$  be possible real root of equation  $f(x)=0$  we know that $a\neq 0$ and$$f'(a)=-\frac{a^5}{120} \neq 0$$