1. ## distance

find the distance from the origin to the point R(m^2-n^2 , 2mn) in simplest form

i got to m^2+2mn-n^2
but i think thats wrong

the answer is m^2 + n^2

2. $OR=\sqrt{(m^2-n^2)^2+4m^2n^2}=\sqrt{m^4+n^4+2m^2n^2}=$

$=\sqrt{(m^2+n^2)^2}=m^2+n^2$

3. I'm a little late but, this picture might help you as well :
Apply the Pythagorean theorem as red dog did and you get the same answer.

4. how do you go from 4m^2n^2 to then having 2m^2n^2??

5. I took red dog's work and expanded it for you. Credits go to him, it's just a few more steps so you can see what's going on.

$\sqrt{(m^2-n^2)^2+4m^2n^2}$

$=\sqrt{(m^2 - n^2)(m^2 - n^2) + 4m^2n^2}$

$=\sqrt{m^4 -2m^2n^2 + n^4 + 4m^2n^2}$

$=\sqrt{m^4+n^4+2m^2n^2}$

$=\sqrt{(m^2+n^2)^2}=m^2+n^2$

6. Originally Posted by eXist
I took red dog's work and expanded it for you. Credits go to him, it's just a few more steps so you can see what's going on.

$\sqrt{(m^2-n^2)^2+4m^2n^2}$

$=\sqrt{(m^2 - n^2)(m^2 - n^2) + 4m^2n^2}$

$=\sqrt{m^4 -2m^2n^2 + n^4 + 4m^2n^2}$

$=\sqrt{m^4+n^4+2m^2n^2}$

$=\sqrt{(m^2+n^2)^2}=m^2+n^2$
thanks heaps
i get most of it now
the only thing is how does the 2m^2n^2 go away??

7. Well you can combine $-2m^2n^2$ with $4m^2n^2$ since both terms have the same variables to the same power. There for you add the coefficients:

$(-2 + 4)m^2n^2 = 2m^2n^2$

From there we have:

$\sqrt{m^4 + n^4 + 2m^2n^2}$

All we are doing from here is factoring that into:

$\sqrt{(m^2 + n^2)^2}$

You can also use substitution if you can't see it right away. Ex: Let $m^2 = a$ $n^2 = b$

That gets you:

$\sqrt{a^2 + 2ab + b^2} = \sqrt{(a + b)^2}$ then resubstitute m and n back in: $\sqrt{(m^2 + n^2)^2}$

Either way, you come up with the same answer.

8. ok i get it now
thanks heaps