## square root of (2x+20) + the square root of (2x)=10

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### square root of (2x+20) + the square root of (2x)=10

This is my problem:
The square root of (2x+20) + the square root of (2x)=10
This is how I worked it: I squared both sides to get 2x+20+2x=10
4x=80
x=20
The answer in the book is 8. I plugged in both answers and 8 is correct. But I don't understand how?
AmySaunders

Posts: 34
Joined: Thu Aug 20, 2009 6:27 pm

### Re: square root of (2x+20) + the square root of (2x)=10

AmySaunders wrote:The square root of (2x+20) + the square root of (2x)=10
This is how I worked it: I squared both sides to get 2x+20+2x=10

How did you get this? Squaring the RHS gives 100, not 10. (10^2 does not equal 10) And what were your steps for the LHS?

$\left(\sqrt{2x+20}+\sqrt{2x}\right)^2$
$\left(\sqrt{2x+20}\right)^2+2*\sqrt{2x+20}\sqrt{2x}+\left(\sqrt{2x}\right)^2$

What then?
Stranger_1973

Posts: 21
Joined: Sun Feb 22, 2009 9:56 pm

### Re: square root of (2x+20) + the square root of (2x)=10

Sorry. I wrote that wrong, the RHS is 100.

I didn't square the whole left hand side, I separately squared each term, if that makes sense.

So, after squaring, I had 2x+20+2x=100

I can see that you squared the whole LHS. Why? In my mind it seems like squaring each term separately should yield the same results.

Thank you for your help, by the way!

Okay, if I square the whole LHS, I FOIL the terms, simplify and end up with:

8x+4square root(10x)=80

but I still have that nasty square root sign. Which is why I squared everything to begin with. To get rid of the square roots so I could solve for x nicely.
AmySaunders

Posts: 34
Joined: Thu Aug 20, 2009 6:27 pm

### Re: square root of (2x+20) + the square root of (2x)=10

Oh, I figured it out. I have to isolate the square root and square both sides, and the answer is 8!

And I understand why you have to square all terms on a side together, and not each separately. It just isn't intuitive to me yet.