Find the minimum number of roots of the equation :
$$\large{p(x)= x^10 + 24x^8 -35x^7+5x^6+25x^5+3x^3+4=0}$$

Question

Find the minimum number of roots of the equation : 
$$\large{p(x)= x^10 + 24x^8 -35x^7+5x^6+25x^5+3x^3+4=0}$$

mathslover · Answer

Which are imaginary ...

maheshmeghwal9 · Answer

Is the first variable$$\LARGE \color{red}{x^{10}}$$?

mathslover · Answer

yes sorry i forgot to put that in brackets

mathslover · Answer

$$\large{\color{violetred}{p(x)=x^{10}+24x^8−35x^7+5x^6+25x^5+3x^3+4=0}}$$

maheshmeghwal9 · Answer

I have done questions like this
but I have forgotten that:(
Can u remind me what concept this question is using?

anonymous · Answer

the mane theory of algebra states that the number of roots of a polinomial is equal to it's degree. In this case degree is 10, so 10 roots

mathslover · Answer

@myko  it is asked for imaginary roots

mathslover · Answer

i just only know about replacing x by (-x) in the equation p(x)

mathslover · Answer

attachment

mathslover · Answer

it is written that if x is replaced by (-x) then the max. number of negative roots for the given equation is given by number of changes in the sign of the equation for -ve roots

mathslover · Answer

@maheshmeghwal9  i only know this theory

mathslover · Answer

can u  give me any link that where i can learn this ..

anonymous · Answer

http://www.purplemath.com/modules/drofsign.htm

mathslover · Answer

from that also i cant get the answer...

maheshmeghwal9 · Answer

If u don't mind I will post the answer after 2 or 2.5c hrs:)

maheshmeghwal9 · Answer

100%

maheshmeghwal9 · Answer

& i hope u gt the answer before i give u:)

mathslover · Answer

ok .. no problem hope for best..

mathslover · Answer

i got the solution is it correct can any 1 help me in this : 

$$\large{p(x)=x^{10}+24x^8-35x^7+5x^6+25x^5+3x^3+4=0}$$

mathslover · Answer

wait..

mathslover · Answer

attachment

mathslover · Answer

is this answer correct ?

mathslover · Answer

@maheshmeghwal9  and @ganeshie8 , @shubhamsrg ...

shubhamsrg · Answer

nops..the actual ans is it has 10 imaginary roots..(source : wolframalpha)
hmmn
and the theorem you used works only when you know that roots are real,,just need to know there sign,,

mathslover · Answer

can u give me the link of that question in wolfram alpha ? @shubhamsrg

shubhamsrg · Answer

http://www.wolframalpha.com/input/?i=x%5E10%2B24x%5E8%E2%88%9235x%5E7%2B5x%5E6%2B25x%5E5%2B3x%5E3%2B4%3D0

mathslover · Answer

i had calculated min. possible imaginary roots..

mathslover · Answer

Minimum number of imaginary roots : 10-4 = 6

shubhamsrg · Answer

what do you mean by min no. of possible roots?

shubhamsrg · Answer

ohh..i understand now..
hmm,,yes,,your sol might be correct @mathslover

maheshmeghwal9 · Answer

@mathslover I have found the way of finding the minimum number of possible imaginary roots of this polynomial.
But I m having problem in taking second, third derivatives.

mathslover · Answer

@maheshmeghwal9 there is no need of deriviatives here ... http://assets.openstudy.com/updates/attachments/4fddb20ee4b0f2662fd2ff1a-mathslover-1339932263149-roots.png

maheshmeghwal9 · Answer

no I m doing this question by STURM's Theorem so wanna take derivatives. @mathslover

mathslover · Answer

but why r u making this more complicated .... well i know that may be important for IIT ...

mathslover · Answer

where r u having the problems in deriviatives post them

maheshmeghwal9 · Answer

k! thats ur wish.
but sturm's theorem is simple dude.
k bye:)

mathslover · Answer

bbye ...

maheshmeghwal9 · Answer

ur answer seems to be right by sturm's theorem too:)

mathslover · Answer

:)

mathslover · Answer

thanks

maheshmeghwal9 · Answer

yw:)