if (SQRT (3-2x))= (sqrt (2x)) + 1, then 4x^2 = ?

a) 1
b) 4
c) 2-2x
d) 4x-2
e) 6x-1

Can you tell me where I am going wrong here?

I square both sides and get:

3-2x = ((sqrt ( 2x ) +1) ^2

-2x = ( ( (sqrt ( 2x ) +1) ^2 ) - 3

square both sides again

4x^2 = [( ( (sqrt ( 2x ) +1) ^2 ) - 3 ] ^ 2

Instead of solving further, I plugged in numbers to test.

for example if x= 2, then 4x^2 = 36 (use equation above).

I would look at the answer choices and plug in x=2. Any answer choice that gave me 36 would be valid.

if x=2
then 4x^2 would be:
a) 1
b) 4
c) 2-2(2)= -2
d) 4(2)-2 = 6
e) 6(2)-1 =11

None of the answer choices are = 36!



HOWEVER, the answer is E. I know the other method to arrive to the answer, I want to know why my reasoning above is incorrect please.

I'm a little lost on the specific notation here.
Will you send me a screen shot of the problem? (help@veritasprep.com)

Veritas Help

VP_Help,
Can you post the answer to this in the thread, I am curious to see how this is done and what the notation is for this problem.

thanks,
John

I'm looking back at your explanation, and I think it comes down to notation, which I'm going to have to see specifically, but I'm guessing it's going wrong in this step:
for example if x= 2, then 4x^2 = 36 (use equation above).

4x^2 isn't 36 here -- x^2 is 4, and 4*4 is 16....

Though again, I think a screen shot is probably our best bet to find a way through this one.

Veritas Help

VP_help,

I didn't ask the question, I was hoping that speedway would have sent you a screen shot. I am curious to see this problem solved out with the proper notation from the beginning.

John R