Question

Square root help, pls?

Answer 1

\[\sqrt{2x^2}+2x=2\]

Answer 2

what do you need to do? solve it..?

Answer 3

(2x^2)^1/2=(2-2x)
square both sides
2x^2=(2-2x)^2
2x^2=4+4x^2-2*2*2x
2x^2=4+4x^2-8x
0=4x^2-2x^2-8x+4
2x^2-8x+4=0
take 2 common
x^2-4x+2=0

Answer 4

find its factor now

Answer 5

@campbell_st
I want to know what x is.

Answer 6

I think I've got it, but it could be wrong,

\[\sqrt{2x ^{2}} +2x = 2\]
\[x \sqrt{2} +2x = 2\]
Now factorise (take out the common factor, which is x)
\[x(\sqrt{2} + 2) = 2\]
Now take the part in brackets over to the other side to get x by itself.
\[x = 2 - (\sqrt{2} - 2)\]
careful of the minus
\[x = 2 - \sqrt{2} - 2\]
\[x = \sqrt{2}\]
hopefully that's right :)

Answer 7

ok... you have 2 choices...

1. complete the square

2. general quadratic formula

Answer 8

use perfect square metohd to find x

Answer 9

x^2-4x+2=0
make a perfect square

Answer 10

@rhiannon1 that doesn't work

Answer 11

Didn't think i was right ... oh well

Answer 12

both methods result in the same answer, its a case of which you're more comfortable with.

Answer 13

and if you wait long enough... someone always gives the answer

Answer 14

x^2-2*2*x+2^2-2^2+2=0
x^2-2*2*x+2^2=(x-2)^2=perfect square
(x-2)^2-2^2+2=0
(x-2)^2-4+2=0
(x-2)^2-2=0
(x_2)^2=2
take root both side
x-2=root 2
x=2+root 2

Answer 15

see, there you go.. an answer...

Answer 16

@ gorv I still don't see how my question got to (2x^2)^1/2=(2-2x)

Answer 17

\[\sqrt{2x^2}+2x=2\]subtract 2x from both sides\[\sqrt{2x^2}=2-2x\]Now rewrite left side as a fractional exponent:\[(2x^2)^{1/2} = (2-2x)\]

\[\sqrt{x} = x^{1/2}\]

Answer 18

i just moved 2x on other side ..and it will become negative on moving on other side

Answer 19

ok.. the 1st step in solving is to subtract 2x from both sides of the equation

which becomes

\[\sqrt{2x^2} = 2 - 2x\]

next step is to square bothsides of the equation.

The right hand side is a binomial so you have

\[(\sqrt{2x^2})^2 = (2 - 2x)^2\]

hope it helps

Answer 20

Ah! I saw \[\frac{(2x^2)^1}{2}=(2-2x)\]
I though gorn was nuts

Answer 21

yeah, he should have put parentheses around the 1/2

what he wrote is what you saw, by operator precedence

Answer 22

or u can write
\[\sqrt{2x^{2}}=\sqrt{2}*\sqrt{x^2}=\sqrt{2}*x\]

Answer 23

only if you know or assume that x>0

Answer 24

it is greater than 0

Answer 25

\[\sqrt{2} *x+2x=2\]

Answer 26

\[x*(\sqrt{2}+2)=2\]

Answer 27

\[x=2\pm\sqrt{2}\]
thanks

Answer 28

No, that's not true — now you need to check for extraneous solutions...

Answer 29

Yeah I can measure it here, it's 3.4 or 0.6
All good thanks!

Answer 30

Try them both out in the original equation.

Answer 31

\[\sqrt{(2+\sqrt{2})^2} + 2(2+\sqrt{2}) = 2\]You really think that's going to be true?!?

Even if we completely ignore the \(\sqrt{2}\) terms (which only make the left side larger), that's still\[\sqrt{2^2} + 2(2) = 2\]That isn't true in my experience...

Answer 32

Sorry, that should be \[\sqrt{2(2+\sqrt{2})^2} + 2(2+\sqrt{2}) = 2\]Even less likely :-)

Answer 33

I'm just kidding, I don't need the answer, I just need to find out how to get square roots unstuck, in this kind of case.
I'm sorry to make you waste you time on it @whpalmer4

Answer 34

In general, if you start out with a radical sign on one side of the equation, and during the process of solving, you square both sides, you have to check for the introduction of extraneous solutions, such as this one.

Answer 35

I'm only wasting my time if you don't pay attention to this important point.

Answer 36

https://www.khanacademy.org/math/algebra/exponent-equations/radical_equations/v/extraneous-solutions-to-radical-equations

Answer 37

This looks good, I'l watch it now. Thanks @whpalmer4