# How do you solve this inequation: √x -3 >= 1/(x - 3)?

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### 3 Answers

We need to solve the inequation: sqrt (x -3) >= 1/(x - 3)

Now square both the sides

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

multiply both the sides with (x -3)^2. Now (x-3)^2 is a positive term and multiplying the two sides of an inequation by a positive term does not change the sign between the two sides.

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

Take the cube root of both the sides

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

=> (x - 3) >=1

Add 3 to both the sides

=> x - 3 + 3 >= 1+3

=> x >= 4

**Therefore x >= 4.**

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

x - 3> =0

x>=3

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

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

Now, we'll subtract 1/(x - 3)^2 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 get 2 cases to analyze:

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 inequality is the interval [4, +infinity).**

sqrt (x -3) >= 1/(x-3).

The inequality is valid only if x> = 3.

So x-3 bieng positive we multiply by (x-3).

(x-3)sqrt(x-3) >= 1.

We square both sides:

(x-3)^3 > = 1.

We take cube root.

x-3 > = 1

x >= 4.

Therefore x >= 1.

(x-3)