What are the real solutions of irrational equation square root(3x+5) + square root(6x-5) = 3*square root x

Since the strategy of solving this type of equations
implies the action of raising the expressions to powers, this could lead to additional
solutions. For this reason, either we can start by imposing constraints of existence of
the square roots, or we can check if the found values are the proper solutions, in the
end.

We'll choose the 1st method and we'll impose
constraints for radicands:

3x + 5 >=0 =>
x>=-5/3 => [-5/3 , +infinite)

6x-5 >=0
=> x >= 5/6 => [5/6 ; +infinte)

x
>= 0 => [0 ; +infinite)

The common interval
of values that make all the square roots to exist is [5/6 ;
+infinte).

Now, we'll raise to square both sides, to remove
the radicals:

[sqrt(3x+5)+sqrt(6x-5)]^2 =
(3sqrtx)^2

3x + 5 + 6x - 5 + 2sqrt(3x+5)(6x-5) =
9x

We'll eliminate like
terms:

9x + 2sqrt(3x+5)(6x-5) =
9x

We'll subtract 9x both sides, to isolate the
radical:

2sqrt(3x+5)(6x-5) =
0

We'll divide by 2 and we'll raise to square again, to
remove the radical:

(3x+5)(6x-5) =
0

We'll cancel each factor of the
product:

3x + 5 = 0 => x =
-5/3

6x - 5 = 0

x =
5/6

Since the negative value doesn't belong to the interval
[5/6 ; +infinte), we'll reject it.

The only
real admissisble solution of the equation is x =
5/6.