What are the solutions of equations square root (x^2-8x+31)-squareroot (x^2-8x+24)=1?

Let [sqrt(x^2 - 8x + 31) - sqrt(x^2 - 8x + 24)] = 1
(1)

We'll multiply the given expression by it's
conjugate:

[sqrt(x^2 - 8x + 31) - sqrt(x^2 - 8x +
24)][sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x + 24)] = [sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x +
24)]

The product from the left returns the difference of
two squares:

x^2 - 8x + 31 - x^2 + 8x - 24 = [sqrt(x^2 - 8x
+ 31) + sqrt(x^2 - 8x + 24)]

We'll eliminate like
terms:

7 = [sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x + 24)]
(2)

We'll add (1) +
(2):

[sqrt(x^2 - 8x + 31) - sqrt(x^2 - 8x + 24)] +
[sqrt(x^2 - 8x + 31) + sqrt(x^2 - 8x + 24)] = 1 + 7

We'll
eliminate like terms:

2sqrt(x^2 - 8x + 31) =
8

We'll divide by 2:

sqrt(x^2
- 8x + 31) = 4

We'll raise to square both
sides:

(x^2 - 8x + 31) =
16

We'll subtract 16 both
sides:

x^2 - 8x + 31 - 16 =
0

x^2 - 8x + 15 = 0

We'll
apply quadratic formula:

x1 = [8+sqrt(64 -
60)]/2

x1 = (8+2)/2

x1 =
5

x2 = (8-2)/2

x2 =
3

Since there is no need to impose
constraints of existence of square roots, because both radicands are positive for any
value of x, therefore the solutions of the equation are: x1 = 5 and x2 =
3.