Graphs.

[stonecold]

so first one dom is [-6,6]
and the second dom is [-4,4]

[TrueTears]

ahh you've confused me cindy!

is it basically that you can't take the square root of a negative number?

u can

[Gloamglozer]

ahh you've confused me cindy!

is it basically that you can't take the square root of a negative number?

u can

But in the world of Methods, the square root of a negative number doesn't exist. 

[cindyy]

if i was to solve for the domain of that
i would have ended up with
x > ±4
thats not right is it?

[cindyy]

ahh you've confused me cindy!

is it basically that you can't take the square root of a negative number?

u can

But in the world of Methods, the square root of a negative number doesn't exist. 

thats what i thought too :S

[the.watchman]

NONONO



AND ,

OR AND , no sols.

[the.watchman]

ahh you've confused me cindy!

is it basically that you can't take the square root of a negative number?

u can

But in the world of Methods, the square root of a negative number doesn't exist. 

thats what i thought too :S

In Methods, that is the case.
But in GMA/Spesh, you learn about complex numbers and the mysterious power of !!!
And all is well!

[TrueTears]

i answered generally to stonecold's question.

[stonecold]

o0o0o i get it.  it will only be a semi circle if the domain is continuous.  if it is the union of two separate domains, then no semi circle, so for dom x> 4 or x<-4, there is no semicircle.

[cindyy]

i also get it now
life is great.

[stonecold]

hehe, i actually learnt something too

[the.watchman]

o0o0o i get it.  it will only be a semi circle if the domain is continuous.  if it is the union of two separate domains, then no semi circle, so for dom x> 4 or x<-4, there is no semicircle.

Yep!

i also get it now Cheesy Cheesy
life is great.

Good, soz for the shocking explanation!

[cindyy]

nah, it was great i was sitting here thinking about it for ages, but finally understand:D

[TrueTears]

the reason it's not a semicircle is because let's let y^2 = x^2-16

now let's assume it is an ellipse (a circle is just a type of ellipse).

can the equation be a circle?

consider focus F_1 and focus F_2

let F_1 (-c,0) and F_2 (c,0)

from the definition of an ellipse we get (a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)

can we get our a and c's? nope

thus it can not be an ellipse thus it can not be a circle and thus it can't be a semicircle.

[dodgedanpei]

so first one dom is [-6,6]
and the second dom is [-4,4]

umm
isn't the first one
(-infinity, -6] U [6,infinity)