# √(x - 3) ≥ 1/(x - 3) help on how to solve please!!!

justaguide | Certified Educator

You want to solve sqrt (x-3) >= 1/ (x-3).

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

Take the square of both the sides.

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

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

Now (x-3)^3 is greater than or equal to 1 if x-3 >= 1

=> x >= 1 + 3

=> x >= 4

Also, as (x-3) is within the square root on the left and in the denominator x-3 on the right, x should be greater or equal to 3.

As any value of x >= 4 satisfies x>=3, we have that x should be greater than or equal to 4.

The required values of x are greater than or equal to 4.

giorgiana1976 | Student

Before solving the inequality, we'll impose constraints of existence of square root.

x - 3 >=0

x >= 3

The interval of admissible values of x is [3 , +infinite)

We'll raise to square both sides, to get rid of the square root:

(x - 3) ≥ 1/(x - 3)^2

We'll subtract 1/(x - 3)^2 both sides:

(x - 3) - 1/(x - 3)^2 ≥  0

We'll multiply by (x - 3)^2 both sides:

﻿﻿(x-3)^3 - 1 >= 0

We'll apply the formula of difference of cubes:

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

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

We'll square raise and we'll combine like terms:

﻿﻿﻿﻿(x-3)^3 - 1 = (x-4)(x^2 - 5x + 7)

(x-4)(x^2 - 5x + 7) >=0

A product is positive if and only if the 2 factors are both negative or both positive.

Case 1)

x-4>0

x>4

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 value of x.

From both inequalities, the interval for admissible values for x is (4; + infinite).

Case 2)

x-4<0

x<4

x^2 - 5x + 7 < 0

But the expression is always positive, for any value of x, so x belongs to empty set.

neela | Student

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

When x< 3, sqrt(x-3) is imaginary .

So x >= 3, then only sqrt(x-3) is real.

When x>3,we multiply both sides by (x-3):

(x-3)^3/2 >=1, if x>3.

We square both sides

(x-3) ^3 >= 1.

We take the cube root of both sides.

x -3 >= 1.

x >= 1+3 = 4.

Therefore the inequlity is valid for all x >=1.