Question

Please solve step by step
I got A
http://awesomescreenshot.com/0832bvu36d

Answer 1

\[\sqrt{2x+4} = 2 + \sqrt x\]
Square on both sides

Answer 2

underroot ( (2x + 4) - x = 2 , 4 = 2x +4 - x , x = 0 .

Answer 3

Sorry... what's going on here? o.o

Answer 4

Square the equation...& tell me what you get..

Answer 5

Also,
\[(a+b)^2 = a^2 + b^2 +2ab\]
I have no idea what Imad did..

Answer 6

I'm a little confused... because there aren't any exponents... Do I have to sq 4 and make it 2^2?

Answer 7

See this -
\[\sqrt{2x+4} - \sqrt x = 2\]
\[\sqrt{2x+4} = 2 + \sqrt x\]
Squaring on both sides
\[(\sqrt{2x+4})^2 = (2 + \sqrt x)^2\]

Now can you write the next step ?

Answer 8

\[\sqrt{4x^2 + 16} = \sqrt{4x^2}?\]

Answer 9

Er..
\[\sqrt y = y^{1/2}\]
So,
\[(\sqrt y)^2 = (y^{1/2})^2 = y\]

Now can you tell me the value of -
\[(\sqrt{2x+4})^2\]

Answer 10

It's just 2x + 4 ...

Answer 11

I think... omg

Answer 12

>-<

Answer 13

Yep.. :)

Answer 14

Now,
\[(a+b)^2 = a^2 + 2ab + b^2\]

Can you tell me the value of -
\[(2 + \sqrt x)^2\]

Answer 15

is it set up like 2^2= (2)(4)(x) + x^2 or...?

Answer 16

Do you mean
2^2 + (2)(2)(x) + x^2
??

Answer 17

Should I have? ummm the 4 was just 2^2=4 ... was I not supposed to do that yet o.o

Answer 18

In this case
a = 2
b = √x

Hence,
\[(2+\sqrt x)^2 = 2^2 + 2*2*\sqrt x + (\sqrt x)^2\]

Answer 19

Now can you solve
\[(\sqrt{2x+4})^2 = (2+\sqrt x)^2\]

Answer 20

\[2x + 4 = 2 + x

Answer 21

2x + 4 = 2 + x ... *

Answer 22

Er..can you write complete equations; I'm not really good with ... :P

Answer 23

Oh sorry... the dots just signify my uncertainty in my response... I'm done lol
2x + 4 = 2 + x
I'm assuming that means I messed up somewhere >-<

Answer 24

Actually I was expecting... <<uncertainty :P
\[2x+4 = 2^2 + 2*2*\sqrt x + (\sqrt x)^2\]

Answer 25

OMG where did all of that come from > -< and did you just say less uncertainty? o.o lol slick...

Answer 26

I explained you how to square the terms separately; then you only had to equate them again.

Answer 27

Ooooh nevermind... pretend I din't say that /.\ I usually forget the beforehand steps when someone's working with me.... what happens next?

Answer 28

Well working with me for a long period of time actually... short attention span.

Answer 29

Now simplify the equation; and collect the radical sign on one side of = and every thing else on the other side.

Answer 30

Before I even attempt to simplify anything may I ask how would one keep the radical sign on one side of the equal sign while moving everything else to the other?

Answer 31

Focus only on simplifying the equation..forget the other part ;)
I'll explain that later..

Answer 32

I went all over the freaking world... It makes no sense to me /.\
I got
\[2x + 4 = 12 \times \sqrt{x} + x\]

If that's correct I'd subtract 4 from both sides and subtract x from both sides then get

\[x = 8 \times \sqrt{x}\]

Except I know I didn't do it right
and it all looks all types of wrong to me ._. sighhhh

Answer 33

Okay, lemme try
\[2x + 4 = 2^2 + 2*2* \sqrt x +(\sqrt x)^2\]
\[2x+4=4+4 \sqrt x + x\]

We can add or subtract a number on both sides of equal to
We can multiply or divide a (non-zero) number on both sides of equal to

So
\[2x+4=4+4 \sqrt x + x\]
\[2x + 4 - (4 + x) = 4 + 4 \sqrt x + x - (4 + x)\]
\[x = 4 \sqrt x\]
Now square on both sides...please don't let me down XD

Answer 34

Definitely no promises.... letting my own self down ;-; lol... um... \[x ^{3/2} = 4\]

Answer 35

Er..by squaring on both sides, I meant
\[x^2 = (4 \sqrt x)^2\]

Don't divide by x (I'll explain why..later)

Answer 36

;-;
x = 4x

Answer 37

>-< omgeez please say yesssssh

Answer 38

Actually;
\[x^2 = 16x\]
Next, we bring everything on the same side of equal to sign
\[x^2 - 16x = 16x - 16x\]
\[x^2 - 16x = 0\]
Then we factorize it (see that x is a common factor in both the terms)
\[x(x - 16) = 0\]

Can you now answer the (original) question ??

Answer 39

z=0 and x=16 ...

Answer 40

z ??

Answer 41

x*

Answer 42

sorries

Answer 43

j/k
Yea..it's correct. :)

Answer 44

Omgeez yay thank you...! Sorry for my incapability to grasp all of these mathematical concepts and steps. Took forever and a day... I appreciate your patience. ^-^

Answer 45

You're welcome :)