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

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

√(x-4) = 1/(x-4)

x-4 =1

x=5

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, to eliminate the square root:

(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 cancel each factor:

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 concerning the nature of roots, we'll accept them.

The equation to be solved is (x-4)^1/2=1/(x-4)

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

square both the sides

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

=> (x - 4)^3 = 1

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

=> (1 - (x - 4))(1 + x - 4 + (x - 4)^2) = 0

=> (1 - x + 4)(1 + x - 4 + x^2 + 16 - 8x) = 0

=> (-x + 5)(x^2 - 7x + 13) = 0

-x + 5 = 0

=> x = 5

x^2 - 7x + 13 = 0

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

=> x1 = 7/2 + i*(sqrt 3)/2

x2 = 7/2 - i*(sqrt 3/2)

**The equation has the solutions x = (5, 7/2 + (sqrt 3)/2, 7/2 - (sqrt 3/2))**