Solve the equation square root(x^2-5)-square root(x^2-8)=1

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

x^2-5>=0

the
expression is positive if x belongs to the
ranges:

(-infinite ; -sqrt5]U[sqrt5 ;
+infinite)

x^2-8>=0

the
expression is positive if x belongs to the
ranges:

(-infinite ; -sqrt8]U[sqrt8 ;
+infinite)

The common intervals of admissible values for x
are:

(-infinite ; -sqrt8]U[sqrt8 ;
+infinite)

Now, we'll solve the equation. We'll move
-sqrt(x^2-8) to the righ side:

sqrt(x^2-5) = sqrt(x^2-8) +
1

We'll raise to square both sides, to eliminate the square
root from the left side:

x^2 - 5 = x^2 - 8 + 1 +
2sqrt(x^2-8)

We'll eliminate x^2 both
sides:

7 - 5 = 2sqrt(x^2-8)

2
= 2sqrt(x^2-8)

We'll divide by
2:

sqrt(x^2-8) = 1

We'll raise
to square again:

x^2 - 8 =
1

x^2 = 9

x1 = 3 and x2 =
-3

Since both values belong to the intervals
of admissible values, the solutions of the equation are: {-3 ;
3}.