Question

I really do not understand how to solve this...can anyone please help?!

Answer 1

check if the solution is extraneous.\[\sqrt{x+\sqrt{2x}}=\sqrt{2x}\]

Answer 2

the whole idea of solving a root function is to get rid of the root.

Answer 3

But how do I do that?

Answer 4

squaring both sids

Answer 5

I'm soo confused..

Answer 6

1st WRITE DOWN that x > 0

If \(\sqrt{a} = \sqrt{b}\), does it make sense that a = b?

Answer 7

Yes, that makes sense!

Answer 8

That's all that is being suggested.

If \(\sqrt{x + \sqrt{2x}} = \sqrt{2x}\), then \(x + \sqrt{2x} = 2x\). Same thing.

Now what?

Answer 9

square both sides? I'm not too sure..:(

Answer 10

You can call it "square both sides" if you like. ALL that happened was exactly like that a and b example to which you agreed. Essentially, if two things are the same, under radicals, they are the same without the radicals. I just removed the top radical from each side.

Are you buying it, yet?

Answer 11

Ohhh okay, that makes sense!

Answer 12

"Square both sides doesn't do any good unless SOME radical is ALONE.

From here: \(x + \sqrt{2x} = 2x\), you must make that radical a lonely fellow before proceeding. Subtract x from each side.

BTW, I was slightly off in my first post. Don't write \(x > 0\), write \(x \ge 0\), instead. I forgot to include x = 0 in the Domain.

Answer 13

So would it be 2x-x=2x-x?

Answer 14

\(\sqrt{2x} = x\) -- Nothing happened to the radical portion.

Answer 15

Oh..

Answer 16

Where does that leave us?

Answer 17

Don't we have to..do something with the x's? get rid of one?

Answer 18

Yes. Use that square root idea again. What do you get?

Answer 19

I got x=0 and x=2

Answer 20

Wow! How did you manage BOTH of them? I expect most folks would miss one.

Did you check them in the ORIGINAL equation?

Answer 21

I have not checked them yet...and squaring roots are easier for me.

Answer 22

Good work. You might just have it!!

Answer 23

Thank you! I'm actually starting to understand it..but how do I know if the solution is extraneous?

Answer 24

Remember that thing I wrote down? \(x \ge 0\) - That's your first clue. If you managed anything less than zero, you would discard it as extraneous. That's the best way to take a stab at it.

The second thing to do would simply be to try it in the ORIGINAL equation.

\(\sqrt{x + \sqrt{2x}} = \sqrt{2x}\)

Substitute x = 2

\(\sqrt{2 + \sqrt{2(2)}} = \sqrt{2(2)}\)

And simplify

\(\sqrt{2 + \sqrt{4}} = \sqrt{4}\)

\(\sqrt{2 + 2} = 2\)

\(\sqrt{4} = 2\)

\(2 = 2\)

Looks like it works.

Do the same with any other PROPOSED solution.