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