# Solve the inequality:square root(x - 3) ≥ 1/(x - 3).

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

x - 3 >=0

We'll add 3 both sides:

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.