How to find the values that $x$ can take to be a real number?

Question

$$3 \cdot \sqrt{x+4} + 5 \cdot \sqrt [8]{6-x}$$ - How to find the values that $x$ can take to be a real number?

I'm a bit confused. However, I want to show my thinkings:

$$ x + 4 > 0 \implies x > -4$$

and

$$6-x>0 \implies 6 >x$$

My Kindest Regards,

Answer 1

You are almost correct! You just haven't considered the case that the radicals could be $0$.

We have $$\sqrt{x+4}=0\implies x=-4$$ and $$\sqrt[8]{6-x}=0\implies 6=x$$

Hence your inequalities become $$x\ge-4,\,x\le6$$ or more succinctly $$-4\le x\le6$$

Answer 2

You should set

$$ x + 4 \ge 0 \implies x \ge -4$$

and

$$6-x\ge0 \implies 6 \le x$$

thus

$$-4\le x\le6$$