Solve the equation `sqrt((6-4x-x^(2)))=x+4`

Explanation

We have `sqrt((6-4x-x^(2)))=x+4`
This equation is equivalent to the system
`{(x+4ge0),(6-4x-x^(2)=(x+4)^(2)):}`
`implies{(xge-4),(x^(2)+6x+5=0):}`
On solving the equation `x^(2)+6x+5=0`
We find that `x_(1)=(-1)` and `x_(2)=(-5)` only `x_(1)=(-1)` satisfies the condition `xge-4`.
Consequently the number `-1` is the only solution of the given equation.
Form 3 An equation in the form
`root(3)(f(x))+root(3)(g(x))=h(x)` .........i
where `f(x),g(x)` are the functions of `x` but `h(x)` is a function of `x` or constant, can be solved as follows cubing both sides of the equation we obtain
`f(x)+g(x)=3root(3)((f(x)(g))(root(3)(f(x))+root(3)(g(x)))=h^(3)(x)`
`impliesf(x)+g(x)+3root(3)(f(x)g(x))(h(x))=h^(3)(x)`
[from Eq. (i)]
We find its roots and then substituting then into the original equation, we choose those which are the roots of the original equation.