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afritts1987
10-23-2008, 07:44 PM
sqrt(x)+sqrt(2x)=1

I tried to solve this by squaring both sides to get rid of the radical over the X. Which left me with

x + sqrt(2x)= 1
then sqrt(2x)= 1-x

I then squared both sides to get rid of the square root....

2x = (1-x)^2
then I got...
2x= x^2-2x+1
and eventually
x^2-4x+1.
I then used the quadratic equation but i keep getting the wrong answer. Did I do this wrong?

mmm4444bot
10-23-2008, 08:13 PM
sqrt(x)+sqrt(2x)=1

I tried to solve this by squaring both sides to get rid of the radical over the X. Which left me with

x + sqrt(2x)= 1

You squared both sides, and all that happened is that one of the radical signs disappeared? That's not correct.

Use FOIL (if you want to take that approach)

MY EDIT: added parenthetical comment to align my response with Subhotash's

Loren
10-23-2008, 08:14 PM
The only thing I see as being wrong is that your line that says x^2-4x+1 should say x^2-4x+1=0. That's what I got. Not seeing your answer, it's hard to say that you are wrong or right. Possibly you made a mistake in applying the quadratic formula or your answer needs further simplification or the answer in the answer book is wrong???

Subhotosh Khan
10-23-2008, 10:21 PM
sqrt(x)+sqrt(2x)=1

I tried to solve this by squaring both sides to get rid of the radical over the X. Which left me with

x + sqrt(2x)= 1
then sqrt(2x)= 1-x

I then squared both sides to get rid of the square root....

2x = (1-x)^2
then I got...
2x= x^2-2x+1
and eventually
x^2-4x+1.
I then used the quadratic equation but i keep getting the wrong answer. Did I do this wrong?

Don't go blindly into the night - squaring terms left and right and hoping to bump into the answer.

Look at your problem - carefully - it is actually very simple.....

\sqrt{x} \, + \, \sqrt{2x} \, = \, 1

\sqrt{x} \cdot (1 + \, \sqrt{2}) \, = \, 1

\sqrt{x} \, = \, \frac{1}{ 1 + \, \sqrt{2}}

and so on.....