Question 1182855


The nature of the roots of the equation

{{{3x^4 + 4x^3 + x - 1 = 0 }}}

observe sign changes

signs are:
{{{3x^4 + 4x^3 + x - 1 }}}
+       +         +    -

=>  number of sign changes is {{{1}}}
it means => number of positive roots is {{{1}}}

now replace {{{x}}} with {{{-x}}} and observe sign changes

{{{3(-x)^4 + 4(-x)^3 + (-x) - 1 }}}

{{{3x^4 -4x^3 -x - 1}}} 
+       -        -    -

=>  number of sign changes is {{{1}}}

=> number of negative roots of original polynomial is{{{ 1}}}


so,
=> number of positive roots is {{{1}}}
=> number of negative roots of original polynomial is {{{1}}} and
=> number of imaginary roots of original polynomial is: 
{{{n-(p+q)}}} .......where {{{n}}} is number of degree,{{{ p}}} is number of positive roots, {{{q}}} is  number of negative roots

{{{n=4}}},{{{p=1}}},{{{q=1}}}

{{{n-(p+q) =4-(1+1)=2}}}

=> number of imaginary roots is {{{2}}}


then answer is

D. One positive real root, one negative real root and two complex roots