# Solve the inequation (x-3)^1/2 - (x-3)^-1>=0

*print*Print*list*Cite

### 1 Answer

We'll start by imposing the constraints of existence of the square root:

x - 3> =0

x>=3

We'll re-write the 2nd term using the negative power property:

**(x-3)^-1 = **1/(x - 3)

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

(x - 3) - 1/(x - 3)^2 >= 0

We'll multiply by (x-3)^2 the inequality:

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

We'll combine like terms inside brackets:

(x - 4)[(x-3)^2 + x - 3 + 1] >=0

A product is zero if both factors have the same sign. We'll discuss 2 cases:

Case 1)

x-4>=0

x>=4

x^2 - 6x + 9 + x - 2 >=0

x^2 - 5x + 7 >=0

x1 = [5+sqrt(25 - 28)]/2

Since delta = 25-28 = -3<0, the expression

x^2 - 5x + 7 > 0 for any x.

The common solution is the interval [4, +infinity).

Case 2)

x-4=<0

x=<4

x^2 - 5x + 7 <0 impossible, because x^2 - 5x + 7 >0 for any value of x.

**The solution of the inequation is the range [4, +infinity).**