Question

Need help with radical equation. Will give medals!

Answer 1

What do you need this time?

Answer 2

\[\sqrt{x-3}+\sqrt{x}=3\]

Answer 3

The answer is 4. I just don't know how to get there algebraically...

Answer 4

Square both sides to get\[x-3+2\sqrt{x^2-3x}+x=9\]then rearrange to get\[\frac{12-2x}{2}=\sqrt{x^2-3x}\]\[(6-x)^2=x^2-3x\]

Answer 5

Is that good?

Answer 6

Then what?

Answer 7

And (6-x)^2 doesn't equal 12-2x

Answer 8

Sorry, I simplified to 6-x, then squared both sides, in the same step.

Answer 9

\[36-12x+x^2=x^2-3x\]\[36-12x=-3x\]

Answer 10

Ah

Answer 11

Thanks!