SOLUTION: Solve and check for extraneous solutions. 3sqrt(w+1)=6 I came up with w=35/3. I don't think that's correct and I'm not sure what I'm doing wrong

Algebra ->  Square-cubic-other-roots -> SOLUTION: Solve and check for extraneous solutions. 3sqrt(w+1)=6 I came up with w=35/3. I don't think that's correct and I'm not sure what I'm doing wrong      Log On


   

Question 295835: Solve and check for extraneous solutions. 3sqrt(w+1)=6
I came up with w=35/3. I don't think that's correct and I'm not sure what I'm doing wrong
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve and check for extraneous solutions.
3sqrt(w+1)=6
-----
sqrt(w+1) = 2
---
Square both sides to get:
w+1 = 4
w = 3
----
Check the answer:
3sqrt(3+1) = 6
sqrt(4) = 2
2 = 2
-----
Solution: w = 3
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Cheers,
Stan H.
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Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Well, let's see how correct your answer is:





And after rationalizing the denominator,



Ok, so back to the drawing board.



Square both sides



Multiply both sides by





Check:







Checks.

John