Solve the equation (x-4)^1/2=1/(x-4)

First, we'll impose the constraints of existence of the
square root:

x - 4>
=0

x>=4

Now, we'll
solve the equation by raising to square both sides:

(x - 4)
= 1/(x - 4)^2

Now, we'll subtract 1/(x - 4)^2 both
sides:

(x - 4) - 1/(x - 4)^2 =
0

We'll multiply by (x-4)^2 the
equation:

(x - 4)^3 -
1>=0

We'll solve the difference of cubes using the
formula:

a^3 - b^3 = (a-b)(a^2 + ab +
b^2)

(x - 4)^3 - 1 = (x - 4  -1)[(x-4)^2 + x - 4 +
1]

We'll combine like terms inside
brackets:

(x - 5)[(x-4)^2 + x - 4 + 1] =
0

We'll put each factor as
zero:

x - 5 = 0

We'll add 5
both sides:

x =
5

(x-4)^2 + x - 4 + 1 =
0

We'll expand the square:

x^2
- 8x + 16 + x - 3 = 0

We'll combine like
terms:

x^2 - 7x + 13 = 0

We'll
apply quadratic formula:

x1 = [7 + sqrt(49 -
52)]/2

x1 = (7 +
i*sqrt3)/2

x2 = (7 -
i*sqrt3)/2

The roots of the
given equation are complex numbers. Since there is not any constraint
imposed with regard to the nature of roots, we'll accept them.