Inequation solve the inequation: square root x -3 >= 1/(x - 3)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remove the square root, hence, you need to square the inequality, such that:

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

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

`((x - 3)^3 - 1)/(x - 3)^2 >= 0 => {((x - 3)^2 > 0),((x - 3)^3 - 1 >= 0):}`

You need to solve the inequality (x - 3)^3 - 1 >= 0, hence, you need to attach its equation, such that:

`(x - 3)^3 - 1= 0 => (x - 3 - 1)((x - 3)^2 + (x - 3) + 1) = 0 =>`

`{(x - 4 = 0),((x - 3)^2 + (x - 3) + 1 > 0):} => x = 4`

Since the equation` (x - 4)((x - 3)^2 + (x - 3) + 1) >= 0` , hence `x >= 4` .

Hence, evaluating the inequality yields `x >= 4.`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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).

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