Question

So I was wondering how to do this, 15-alt="\sqrt{x+2}" align="absmiddle" class="latex-formula"> what value of x does the expression make sense

Answer 1

Answer:

x >= -2

Step-by-step explanation:

For real numbers (like, for graphing or applications) the number inside the square root symbol must be a positive number. So, if x is -3, -4, -5 you will get a negative number inside the square root. If x is -2, you get a zero, which is fine(allowed, no problem) If x is -1, you get a 1 inside there, which is no problem. Any positive number for x yields a positive number inside the square root.

So x must be greater than -2.

Saying what x is or can be is what is "the domain"

Domain: {x | x >= -2}

This is set notation.

Interval notation looks like this:

[-2, infinitysymbol)

Simply, x >= -2

Answer 2

Answer:

Step-by-step explanation:

The question makes sense for all values of x, but I imagine that is not exactly what you mean.

I think you mean for what values of x does the question produce a real square root?

The answer is as long as x + 2  does not drop below   0, the results will be real.

Put mathematically x ≥ -2 gives a real result. For now just don't try and take the square root of a minus number.