Question

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

Answer 1

`:. 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`

Answer 2

`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`