For $x$ and $k$ real numbers, for what values of $k$ will the graphs of $f(x)=-2\sqrt{x+1}$ and $g(x)=\sqrt{x-2}+k$ intersect? 
For $x$ and $k$ real numbers, for what values of $k$ will the graphs of $f(x)=-2\sqrt{x+1}$ and $g(x)=\sqrt{x-2}+k$ intersect?

I tried to make an equation of them, but I’m stuck with the two variables and I couldn’t solve it. Much appreciation.
We didn’t do calculus yet..
 A: Rearranging, we have:
$$k = -2 \sqrt{x+1} - \sqrt{x-2}$$
The domain of the RHS is $[2, \infty)$. Therefore, the graphs will intersect when
$$k ≤ -2 \sqrt{2+1} - \sqrt{2-2} \Rightarrow k ≤ -2 \sqrt3.$$
A: You want $-2\sqrt{x+1}=\sqrt{x-2}+k\; (*)$ has at least one $x$-solution.
If we call $h(x)=-2\sqrt{x+1}-\sqrt{x-2}$ then we have that $h$ is defined only for $x\ge 2$ and $h(2)=-2\sqrt 3$ while $\lim_{x\to\infty}h(x)=-\infty$. Then $$h'(x)=-\frac 1{\sqrt{x+1}}-\frac 1{2\sqrt{x-2}}=-\frac{2\sqrt{x-2}+\sqrt{x+1}}{2\sqrt{x+1}\sqrt{x-2}}<0\; \forall x>2$$
So $h$ is strictly decreasing and for intermediate values theorem it is surjective on $(-\infty,-2\sqrt 3]$. We conclude that $(*)$ has exactly one solution for all $k\le -2\sqrt 3$ an no solution for other values.
A: Hint: You must solve the equation
$$\sqrt{x-2}+k=-2\sqrt{x+1}$$ for $$x\geq 2$$ Writing this equation in the form
$$-k-\sqrt{x-2}=2\sqrt{x+1}$$ then it must be
$$k^2+2\geq x$$ and you can square it.
After squaring one times we get
$$2k\sqrt{x-2}=3x+6-k^2$$
Squaring again we get
$$-k^4+10 k^2 x+4 k^2-9 x^2-36 x-36=0$$
Solving this we get
$$k\leq -2 \sqrt{3}\land x=\frac{1}{9} \left(5
   k^2-18\right)-\frac{4}{9} \sqrt{k^4-9 k^2}$$
A: This problem is not about solving an equation in the variables $x$ and $k$. Keep in mind that you're looking for values of $k$ such that there exist $x$ such that $f(x)=g(x)$.
$f$ is decreasing and $g$ is increasing for any value of $k$. edit Furthermore, $\lim_{x \to \infty} f(x) = -\infty$ and $\lim_{x \to \infty} g(x) = +\infty$ . /edit
Since $g$ is defined only for $x\geqslant 2$, you hence get :
$$\Gamma_f \textrm{ and } \Gamma_g \textrm{ intersect } \Longleftrightarrow g(2) \leqslant f(2) $$
But $g(2) = k$. Hence, $\Gamma_f$ and $\Gamma_g$ intersect iff $k\leqslant -2 \sqrt{3}$.
