Finding the domain of $\sqrt{x^2-7}$

Question

How do you find the domain of $\sqrt{x^2-7}$? The answer is $(-\infty ,-\sqrt7)\cup (\sqrt7,\infty)$.

Answer 1

$$\sqrt{x^2-7}$$ is a real number if and only if $$x^2\ge 7$$

For positive values of x, $$x^2\ge 7 \implies x\ge \sqrt 7 $$

and for negative values of x, $$x^2\ge 7 \implies x\le -\sqrt 7 $$

Thus the domain is$$ \left(-\infty,-\sqrt{7}\right]\cup \left[\sqrt{7},\infty\right)$$

Answer 2

We require $x^2-7 = \left(x-\sqrt{7}\right) \left(x+\sqrt{7}\right) \geq 0$. Sketching a graph of $y=x^2-7$ shows us that the regions we want are $x\geq\sqrt{7}$ and $x\leq-\sqrt{7}$.

ie we require $x\leq-7$ which is equivalent to $x\in \left(-\infty,-\sqrt{7}\right]$ OR (which is the same as $\cup$) $x\geq 7$ which is equivalent to $x\in \left[\sqrt{7},\infty\right)$, so we have the domain is $\left(-\infty,-\sqrt{7}\right]\cup \left[\sqrt{7},\infty\right)$.

(NB then answer given was slightly wrong as it did not include the values of $\pm\sqrt{7}$ in the domain — they should be included as they are defined under this function)

Answer 3

$x^2 \geq 7$ (as you cannot have negative numbers under square roots!)

$\implies x=\sqrt{7}$ or $ x=-\sqrt{7}$ are the minimum values $x$ can take.

Hence you have $(-\infty ,-\sqrt7]\cup [\sqrt7,\infty)$ as the answer.

Answer 4

Knowing only basic results on absolute values: $$x^2-7\ge 0\iff x^2\ge 7\iff\sqrt{x^2\strut }=|x|\ge\sqrt 7\iff x\ge \sqrt 7\:\text{ or }\:x\le -\sqrt 7. $$