Question

Can somebody please explain to me how my teacher we from this....
(m^2-n^2/m^2+n^2)^2 + (2mn/m^2+n^2)^2=1
to this
(1-(n/m)^2/1+(n/m)^2)^2 + (2(n/m)/1+(n/m)^2=1

She said something abut dividing top and bottom by m^2, but I just cant figure it out.

Answer 1

\[({m^2-n^2\over m^2+n^2})^2+({2mn\over m^2+n^2})^2=1\]

Answer 2

correct?

Answer 3

yes

Answer 4

\[({m^2-n^2\over m^2+n^2})^2+({2mn\over m^2+n^2})^2=1\]\[\left({1-(\frac nm)^2\over 1+(\frac nm)^2}\right)^2+\left({2(\frac nm)\over 1+(\frac nm)^2}\right)^2=1\]sorry this took so long to type out, but I need to be sure...
so this is what we want to figure out?

Answer 5

yes yes yes

Answer 6

ok, so the first fraction is having top and bottom divided by m^4, and the second is having top and bottom divided by m^2

Answer 7

look at the second term first

Answer 8

sorry, actually I look at it like multiplying both expressions by (1/m^4)/(1/m^4)
here is the second term\[\large ({2mn\over m^2+n^2})^2={\frac1{m^4}\over\frac1{m^4}}({2mn\over m^2+n^2})^2=({\frac{2mn}{m^2}\over\frac{m^2+n^2}{m^2}})=\left({2(\frac nm)\over 1+(\frac nm)^2}\right)^2\]notice that by letting the m^4 into the (...)^2 part means that it becomes m^2 inside the parenthesews

Answer 9

typo, dropped a square on the third part above\[\large ({2mn\over m^2+n^2})^2={\frac1{m^4}\over\frac1{m^4}}({2mn\over m^2+n^2})^2=\left({\frac{2mn}{m^2}\over\frac{m^2+n^2}{m^2}}\right)^2=\left({2(\frac nm)\over 1+(\frac nm)^2}\right)^2\]

Answer 10

1/m^4*m^2+n^2, why doesn't the m^2 cancel?

Answer 11

it does, that's why we are left with 1+(2/m)^2 in the denom

Answer 12

ok I see

Answer 13

look at just in the denom\[\frac1{m^4}(m^2+n^2)^2=\left(\frac1{m^2}(m^2+n^2)\right)=({\cancel{m^2}^1+n^2\over \cancel{m^2}})=1+\frac{n^2}{m^2}=1+(\frac nm)^2\]

Answer 14

dropped the whole squared thing again :/

Answer 15

how did you you to multiply both expressions by 1/m^4???

Answer 16

I multiplied by\[{\frac1{m^4}\over\frac1{m^4}}=1\]so it's legal ;)

Answer 17

I think this is to advance for me, I am really confused.

Answer 18

I just divided top and bottom by the same number; the same way you simplify a fraction
the only tricky part is the fact that we need to be careful about what happened when we divide top and bottom of each fraction by m^4

Answer 19

I mean 1/m^4(m^2+n^2)=(1/m^2(m^2+n^2) how did it go from m^4 to m^2?

Answer 20

look at it the other way

Answer 21

it makes more sense to me just divided both sides by m^4, as you said originally

Answer 22

1/m^2(2mn)=2mn/m^2 this confuses me.

Answer 23

\[\frac1a\cdot b=\frac1a\cdot\frac b1=\frac ba\]oops flipped it, my bad

Answer 24

multiplying by 1/(something) is the same as dividing by (something)

Answer 25

sorry it was 1/m^4(2mn)=2mn/m^2 how did the 1/m^4 cancel down?

Answer 26

because we had to put it in the parentheses
here is an example\[({2n\over m})^2=\frac1{m^2}(2n)^2\]so to pull the m out of the squared parentheses, we had to square it

the same goes the other way; to put the m^2 into the parentheses we would have to take the sqaure root

Answer 27

\[\frac1{m^2}(2n)^2=({2n\over m})^2\]now in your case we have m^4, so to put it into the parentheses we have to take the square root, which is m^2

Answer 28

\[\frac1{m^4}(2nm)^2=({2nm\over m^2})^2\]it went from ^4 to ^2 because we had to put it into the parentheses

Answer 29

I have to go, hopefully someone can continue where I left off if you need more help

Answer 30

Is this an algebraic fraction?

Answer 31

I guess

Answer 32

You have been amazing.

Answer 33

thanks, just keep staring at it and it will make sense I'm sure :)
but feel free to bump your question for more help
later!