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

First we will square both sides:

===> [sqrt(x^2 + 3x -4)]^2 = [ sqrt(x^2 - 2)]^2

==> (x^2 + 3x - 4 = x^2 -2

Now we will combine like terms:

==> x^2 + 3x - 4 - x^2 + 2 = 0

==> x^2 - x^2 + 3x - 4 +2 = 0

==> 3x - 2 = 0

Now add 2 to both sides:

==> 3x = 2

==> x = 3/2

**==> the answer is x = 3/2**

We'll impose the constraints of existence of the square root:

x^2 + 3x - 4>=0

x^2 + 3x - 4 = 0

(x+4)(x-1) = 0

x^2 + 3x - 4 is positive for x belongs to the intervals (-infinite,-4] U [1,+infinite).

x^2 -2>=0

x^2 -2 is positive for x belongs to the intervals (-infinite,-sqrt2] U [sqrt2,+infinite).

The intervals of admissible values of x are: (-infinite,-4] U [sqrt2,+infinite).

We'll square raise both sides:

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

We'll eliminate x^2:

3x - 4 = -2

We'll add 4 both sides:

3x = 2

x = 2/3

**Since x = 2/3 does not belong to the intrvals of admissible values, we'll reject it.**

To solve for x in sqrt(x^2+3x-4) = sqrt(x^2-2):

If sqrta = sqrtb, then a = b.

So sqrt(x^2+3x+4) = sqrt(x^2-2) implies x^2+3x-4 = x^2-2.

3x-4 = -2.

3x = -2+4.

3x = 2.

3x/3 = 2/3.

x = 2/3.

Therefore x = 2/3 is the solution of the given equation.