SOLUTION: I would appreciate it if a tutor could verify my answer to the below question. Solve for x: sqrt(5-x) +3=sqrt(3x+4) My answer: sqrt(3x+4)^2=[(sqrt(5-x)+3]^2 3x+4=(

Algebra ->  Radicals -> SOLUTION: I would appreciate it if a tutor could verify my answer to the below question. Solve for x: sqrt(5-x) +3=sqrt(3x+4) My answer: sqrt(3x+4)^2=[(sqrt(5-x)+3]^2 3x+4=(      Log On


   

Question 6843: I would appreciate it if a tutor could verify my answer to the below question.

Solve for x: sqrt(5-x) +3=sqrt(3x+4)

My answer:

sqrt(3x+4)^2=[(sqrt(5-x)+3]^2
3x+4=(5-x)+(6)sqrt(5-x)+9
3x+4=14-x+(6)sqrt(5-x)
4x-10=(6)sqrt(5-x)
(4x-10)^2=[(6)sqrt(5-x)]^2
16x^2-80x+100=36(5-x)
16x^2-80x+100=180-36x
16x^2-44x-80=0
(16x+20)(x-4)
16x+20=0 or x-4=0
16x/16=20/16 or x=4
x=-5/4 or x=4
****x=-5/4 is an extraneous solution****

In advance, thank you for your assistance!
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
you are correct, but the -5/4 value is a valid answer..if you know Complex numbers. If not, then you can indeed ignore this answer.
Well done.
Jon.