Quadratic Equations: solving sqrt(x) + sqrt(2x) = 1

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afritts1987

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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?
 
Re: Quadratic Equations

afritts1987 said:
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)
Click to expand...


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

 
Re: Quadratic Equations

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???
 
afritts1987 said:
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?
Click to expand...

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.....

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

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

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

and so on.....
 
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