Inequality with square roots where solution found with discriminant is not valid

Question

I have :

$3 + \sqrt{x-1} > \sqrt{2x}$

when doing basic algebraic operations

I will find that :

$ 0 > x^{2} -52x + 100 $

I will use $b^2-4ac$ formula

to eventually find out that there are 2 roots $x_1 = 50 $ and $ x_2 = 2$

However I will find out when testing these two results that $x_2$ is not the solution.

I also know that $ x > 1$

The result should be $ x[1,50) $ but I can not understand how can I interpolate this result from results above.

Is there some fault in my reasoning to find solution with discriminant?

Answer 1

The domain gives $x\geq1$ and we have $$9+6\sqrt{x-1}+x-1>2x$$ or $$6\sqrt{x-1}>x-8.$$ Now, for $x\leq8$ it's true, but for $x>8$ we obtain $$36(x-1)>(x-8)^2$$ or $$x^2-52x+100<0$$ or $$(x-2)(x-50)<0$$ or $$2<x<50,$$ which with $x>8$ gives $$8<x<50$$ and we got the answer: $$[1,50).$$