for what of x will square root of x^2-4 represent a real number??

Question

anonymous · Answer

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anonymous · Answer

x >= 2, to make sure that you don't have a negative result within the radical sign...

anonymous · Answer

if the result within or inside the radical becomes negative then you'll have an imaginary number not a real number...

anonymous · Answer

how??

anonymous · Answer

i dunno how to get it O.O

anonymous · Answer

solving for x, by equating the whole expression to zero and square it....

anonymous · Answer

$$\sqrt{(x^2-4)}\ge0$$$$x^2-4\ge0$$$$x^2\ge4$$
so either $$x \ge2 \quad or \quad  x \le -2$$
because a square is always positive

anonymous · Answer

there you are...

anonymous · Answer

It will always be a positive real result unless the (x^2-4) is less than 0. Which will only happen if -2<x<2

anonymous · Answer

that's more clear than mine... :)

anonymous · Answer

how help please :(

anonymous · Answer

What don't you understand?