Irrational equation with quadratic equation s under square roots

[mathwannabe]

Hello everybody

I got this problem, I did it and I got the right solution, but I want' to know if the method I used to get it is good. Here it is:

1) $\displaystyle \sqrt{y^2+4y+8}+\sqrt{y^2+4y+4}=\sqrt{2(y^2+4y+6)}$

When I saw this, I said to myself "oh man, I'm gonna have 1000 terms here and there will be 3rd an 4th degrees So I said:

$\displaystyle t=y^2+4y+6$

Then I had:

$\displaystyle \sqrt{t+2}+\sqrt{t-2}=\sqrt{2t}$

Then, I set my condition for $\displaystyle t$ to be $\displaystyle t\geq2$

I solved for $\displaystyle t$ and got $\displaystyle t_{1}=2$ and $\displaystyle t_{2}=-2$

I discarded $\displaystyle t_{2}$ because it didn't meet previously set condition and solved $\displaystyle y^2+4y+6=2$ and got that $\displaystyle y_{1}=-2$. That is my solution for this entire problem and it matches with solution provided by the book. Is this workflow correct?

EDIT: I have put this thread here by mistake. This thread should be in the "Beginning Algebra" section.

 

[mathwannabe]

Looks fine to me. When is your exam?

It is around the end of the month... between 25th of June and 5th of July.

 

[mathwannabe]

Looks fine to me. When is your exam?

By the way Jeff, is the word I used ... "condition for t" ... right? I didn't know how to translate so I improvised