How do you solve #3x^4+11x^3=4x^2#?

2 Answers
Mar 27, 2016

Answer:

#:. x=0,0,1/3,-4#

Explanation:

#x^2(3x^2+11x-4)=0#
#=x^2(3x^2+12x-x-4)=0#
#=x^2(3x(x+4)-1(x+4)=0#
#=x^2(3x-1)(x+4)=0#
#:. x=0,0,1/3,-4#

Mar 27, 2016

Answer:

#x=0# or #x=1/3# or #x=-4#

Explanation:

We have #3x^4+11x^3=4x^2# and now let us bring the monomial to LHS, which gives us

#3x^4+11x^3-4x^2=0# and now taking #x^2 common this becomes

#x^2(3x^2+11x-4)=0#

Now let us factorize #3x^2+11x-4#, for which we have to identify two numbes whose sum is #+11# and product is #3xx(-4)=-12#. Clearly these are #+12# and #-1# and splitting middle term, we get

#x^2(3x^2+12x-x-4)=0# or

#x^2(3x(x+4)-1*(x+4))=0# or

#x^2(3x-1)(x+4)=0#

Hence, either #x=0# or #3x-1=0# or #x+4=0#

i.e. #x=0# or #x=1/3# or #x=-4#