Roots of x inside square root

Question

In finding the roots of

$ \sqrt{6-4x-x^2}=x+4$

I get that the roots of $x$ are $-5$ and $-1$. However first I need to take into account that:

• $x+4 \ge 0$ and
• $6-4x-x^2 \ge 0$.

Considering the second equation gives me:

$x \ge -2+ \sqrt{10}$,

would the roots of $x$ I found in the first equation be considered invalid?

Answer 1

It must be $$6-4x-x^2\geq 0$$ and $$x\geq -4$$, solving these inequalities we get $$-4\le x\le -2+\sqrt{10}$$ so we get after squaring and solving the original equation $$x=-1$$

Answer 2

You are right to be concerned, but your work needs correction.

The expression $6-4x-x^2 = 10 - (x + 2)^2$ is non-negative when $(x+2)^2 \leq 10$ i.e. $-2-\sqrt 10 \leq x \leq \sqrt 10 - 2$. This includes both the points $-5$ and $-1$, so you do not face domain issues in this question.

You do face codomain issues, however, since $x+4 \geq 0$ is also required, which leads to $x \geq -4$,thus giving $-1$ as the only solution.

Note that $-5$ is a solution to $\sqrt{6-4x-x^2} = \color{red}{-(x+4)}$, but squaring both sides of this equation leads to the same quadratic equation with roots $-5,-1$, so you should keep in mind the extraneous roots coming from squaring : if you have $f(x) = g(x)$ and wish to square it and find roots, you will also get the roots of $f(x) = -g(x)$.

Answer 3

Let $x+4 \ge 0~~~(1)$ and for tje reality of the roots $$6-4x-x^2 \ge 0~~~~~~(2) \Rightarrow x^2+4x-6 \le 0 \Rightarrow -2-\sqrt{10} \le x \le -2+\sqrt{10}.$$ So that we can square the given equation, then we have $$x^2+6x+5 \Rightarrow x=-5, -1$$. As per the conditiona (1,2), $x=-1$ is the correct root.