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

giorgiana1976 | Student

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.