MHB GRE.al.4 Find the domain of \sqrt{x^2-25}

[karush]

$\tiny{GRE.al.4}$
Find the domain of $f(x)=\sqrt{x^2-25}$
a. $[x\le-5]U[x\ge 5]$
b. x=5
c. $5 \le x$
d. $x\ne 5$
e. $\textit{all reals}$

well just by observation because of the radical I chose c.
but was wondering if imaginary numbers could be part of the domain alto it is not asked for here

https://dl.orangedox.com/QS7cBvdKw55RQUbliE

 

[skeeter]

$x^2-25 \ge 0 \implies x^2 \ge 25 \implies |x| \ge 5$

also, look at the graph of $y = x^2-25$

you want to try another choice?

 

[karush]

oh because x can be $\pm 5$ and outside the graph so a.

 

[HOI]

Since you ask about complex numbers, if all complex numbers are allowed as values for this function then its domain is "all complex numbers".

In order that this question make sense, we must be requiring that the function value, $\sqrt{x^2- 25}$ be a real number. That means that $x^2- 25\ge 0$.

$x^2- 25= (x- 5)(x+ 5)\ge 0$.
The product of two numbers is positive if and only if the two numbers have the same sign- both postive or both negative.

x- 5 and x+ 5 are both positive, x- 5> 0 and x+ 5> 0 so x>5 and x> -5. Of course, if x> 5 then x is also greater than -5 so $x\ge 5$ is a solution.

x- 5 and x+ 5 are both negative, x- 5< 0 and x+ 5< 0 so x< 5 AND x< -5. Of course, if x< -5 then x is also less than 5 so $x\le -5$ is a solution.

The solution set is $\{x| x\le -5 or x\ge 5\}$.