Question

√(2b - a+b)^2 + (2c - c)^2

Answer 1

@LoveYou*69

Answer 2

Oh, there you are...

Answer 3

What are we supposed to do, simplify?

Answer 4

\[\Large \sqrt{(2b - a+b)^2 + (2c -c)^2}\]

Answer 5

I'm supposed to simplify that and √(2a - a+b)^2 + (0 - c)^2 enough that I can tell they are equal. I can do this one if u help me with that one first :P

Answer 6

Of course :)
2b - a + b is just 3b - a, correct?

Answer 7

Yeah, I think, and 2c -c is c so is it √(3b-a)^2 + (c)^2 ??

Answer 8

That is correct...
\[\Large = \sqrt{(3b-a)^2 +(c)^2}\]

Answer 9

Okay thank you.

Answer 10

Now, what about the other one?

Answer 11

I can do the other one lol hold on

Answer 12

\[\Large \sqrt{(2b - a+b)^2 + (0 -c)^2}\]

Answer 13

MU = √(a + b)^2 + (-c)^2 MU = √(a + b)^2 + c ?

Answer 14

Hang on, this one...
\[\Large \sqrt{(2b - a+b)^2 + (2c -c)^2}\]
simplifies into
\[\Large \sqrt{(3b-a)^2 + (c)^2}\]
yes?

Answer 15

yes

Answer 16

and this...
\[\Large \sqrt{(2b - a+b)^2 + (0-c)^2}\] simplifies into\[\Large \sqrt{(3b - a)^2 + (-c)^2}\]
But \[\Large (-c)^2=c^2\]
so...?

Answer 17

(-c)^2 = c^2? why? wouldnt it be just c?

Answer 18

Because it isn't? LOL
Try -4.
\[\Large (-4)^2 =_? \ 4 \qquad \color{red}?\]
I don't think so :P
\[\Large (-4)^2 = \color{blue}{4^2}\]

Answer 19

True ._. I feel blonde.

Answer 20

LOL so this ...
\[\Large \sqrt{(3b - a)^2 + (-c)^2}\]is just \[\Large \sqrt{(3b-a)^2 +c^2}\]
the same as the one earlier, thus showing that they are equal :)

Answer 21

Okay thank youu!

Answer 22

Seee, the only problem is if you go and look at the second equation i gave you it is √(2a - a+b)^2 + (0 - c)^2, which would equal √(a+b)^2 + (c)^2?

Answer 23

@terenzreignz

Answer 24

Yes... what's wrong with that?

Answer 25

Then they aren't equal D:

Answer 26

Why not?

Answer 27

√(a+b)^2 + (c)^2 and √3b - a)^2 + (c)^2

Answer 28

wait, why is it (a+b)^2?
It is also (3b - a) check it again...

Answer 29

√(2a - a+b)^2 + (0 - c)^2 ???? (2a - a+b)^2 equals (a+b)^2 right??

Answer 30

Oh?
Could you post the original question again? Just so we're clear, and we'll start over :D

Answer 31

Prove MD = MU:
Original Q: MD = √(2b - a+b)^2 + (2c - c)^2
MD = √3b - a)^2 + (c)^2
Original Q: MU = √(2a - a+b)^2 + (0 - c)^2
MU = √(a + b)^2 + (-c)^2
MU = √(a + b)^2 + (c)^2

Answer 32

Well, it can't be proved unless that isn't all of it...

Answer 33

D:

Answer 34

Why is it MD and MU anyway?

Answer 35

I have to prove the diagonal is bisecting DU and M is the midpoint so I need the distance from m and d and m and u to see if theyre equal proving that it is being bisected.

Answer 36

May I see the figure?

Answer 37

yeah

Answer 38

https://courses.vlacs.org/vlacontent/geometry_v8_gs//content/module05/imagmod05/05_08d_32.gif

Answer 39

"Prove(5-4): MD=MU and MA=MQ"

Answer 40

LOL this helps :)

Answer 41

.-.

Answer 42

NO, really :D
for one thing, you don't really have to prove that the SQUARE ROOTS of the stuff are equal, you didn't need to bother, you just needed the insides of the square roots to be equal

Answer 43

Okay, so let's get MD^2 instead...
(this just removes the radical)
\[\Large [2b - (b+a)]^2 +(2c-c)^2\]

Answer 44

Do you see your mistake? You forgot to put a+b inside parentheses...

Answer 45

good think I asked you to show me the figure :P

Answer 46

This now simplifies into...
\[\Large (b+a)^2 + c^2\]

Answer 47

And everything's okay now, I hope? Let's check MU

Answer 48

thats the same thing i said tho. wouldnt it be now (b - a)^2? and c^2

Answer 49

Whoops, I meant..
\[\Large (b-a)^2 + c^2\]

Answer 50

Yeah, so that's.
\[\Large MD^2\]

Answer 51

Let's check \(MU^2\)

Answer 52

\[\Large [2a -(a+b)]^2+(0-c)^2\]correct?

Answer 53

yes

Answer 54

Well, this simplifies into...
\[\Large (a-b)^2 + (-c)^2\]

Answer 55

But you know...
\[\Large (a-b)^2 = (b-a)^2\]

and
\[\Large (-c)^2 = c^2\]
so.......

Answer 56

So we have: (b - a)^2 +( c)^2 and (a- b)^2 + (c)^2? whihc are equal?

Answer 57

Yes :)
so consequently, since they are equal, their square roots are equal :)

Answer 58

Or you could work them while already under the square root, it makes no difference :)

Answer 59

lol okay, well now i gotta solve
MA = √((2a+2b) - (a+b))^2 + (2c - c)^2
=
MQ = √(0 - (a+b))^2 + (0 - c)^2
but i can prolly just go do it on my ownn

Answer 60

Well, just ring me if you run into any more problems :)
And for goodness' sake, post the question :P
Imagine how hard this'd be if it was all in chat o.O

Answer 61

:PPP

Answer 62

I cannot believe I haven't fanned you yet -.-