How do you solve #\sqrt{- 2- 2r} = 1- \sqrt{- 3- 4r}#?

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Answer 1 · 2016-09-17T09:16:34

#r = -1#

#\sqrt{- 2- 2r} = 1- \sqrt{- 3- 4r}# so

#\sqrt{- 2- 2r}+\sqrt{- 3- 4r}=1# squaring both sides

#-2-2r+2sqrt(-2-2r)sqrt(-3-4r)-3-4r=1# or

#sqrt(-2-2r)sqrt(-3-4r)=3+3r# squaring again

#2(r+1)(3+4r)=3^2(r+1)^2# and solving for #r#

we get at

#r = -3# and #r = -1#.

The only feasible solution being #r = -1#