Solve for x sqrt(2x+5) = sqrt(x+2) + sqrt(2x-3).

Before solving the equation, we'll set the constraint of existence of the sqrt.

2x + 5 >=0

2x>=-5

x>=-5/2

x+2>=0

x>=-2

2x-3>=0

2x>=3

x>=3/2

The interval of admissible values for x is [3/2 , +inf.)

Now, we'll solve the equation, raising to square both sides, in order to eliminate the square root:

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

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

We'll keep only the sqrt to the right side, the rest of the terms we'll move them to the left side:

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

We'll eliminate like terms:

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

We'll raise to square again:

(6-x)^2 = 4*(x+2)(2x-3)

We'll expand the square and we'll remove the brackets:

36 - 12x + x^2 = 8x^2 + 4x - 24

We'll move all terms to one side:

36 - 12x + x^2 - 8x^2 - 4x + 24 = 0

We'll combine like terms:

-7x^2 - 16x + 60 = 0

We'll multiply by -1:

7x^2 + 16x - 60 = 0

We'll apply the quadratic formula:

x1 = (-16 + 44) / 14

x1 = 2

x2 = (-16 - 44) / 14

x2 = -60 / 14

x2 = -30/7

Since the value of the first root belongs to the interval of admissible values of x, the equation will have only one solution, namely x = 2.

sqrt(2x+5) = sqrt(x+2)+sqrt(2x-3). To solve for x.

We  squaring both sides we get

2x+5 = x+2 + 2sqrt{(x+2)(2x-3)} +2x-3

2x+5  = 3x-1 +2sqrt{x+2)(2x-3)}

2x+5-3x+1 = 2sqrt{(x+2)(2x-3)}

-x+6 = 2sqrt{x+2)(2x-3)}. Again sqre both sides:

(-x+6)^2 = 4(x+2)(2x-3)

x^2-24x+36= 4{2x^2 -3x+4x-6}

x^2-24x+36 = 8x^2 +4x-24

0 = (8-1)x^2 +(24+4)x -24-36

7x^2 +28x-60 = 0

x =  {-28 +or- sqrt(28^2 - - 4*7*60}/2*7

x1 = {-28 + sqrt(2864)}/14.

x2 {-28 -sqrt(2864)}/14