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sqrt{3x-5}=2-sqrt{x-1}
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Square both sides.
\[3x-5=4-4\sqrt{x-1}+x-1\]
find all real and imaginary solutions to equation
\[2x-8=4\sqrt{x-1}\]
Now divide both sides by 2 and then square both sides again.
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\[x-4=2\sqrt{x-1}\]
\[2x-8=-4\sqrt{x-1}\]
\[x^2-8x+16=4(x-1)\]
mertsj did you see kropots answer it would be -4
@kropot72 The negative makes no difference. You're about to square it anyway.
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gotcha ty
If it makes you feel better, we can include it? \[x-4=-2\sqrt{x-1}\]
Now square both sides.
no i see your point
\[x ^{2}=12x-20\]
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Collect all the terms onto the left hand side and you will have a quadratic to solve
x-10 and x-2 are my answers
x = 10 and x = 2 are the solution
ty kropot72 and mertsj
All credit to mertsj.
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