Find the range of values of x from this inequality equation.

Question

|√(X²-8x+12)| ≥ 0

Note:

( is bracket

| is modulus not bracket

Answer 1

( -∞ , 2 ] U [ 6 , ∞ )

The value inside of the modulus needs to be ≥ both of the positive and negative value of the right side of the inequality. Since this value is 0, it cannot be positive or negative, so we will just solve for what makes the value inside of the modulus ≥ 0.

Below I will solve this inequality:

√(x²–8x+12) ≥ 0

(Square both sides to remove the square root)

x²–8x+12 ≥ 0

I will set the inequality equal to 0 and then factor the polynomial to solve for x:

(x²–8x+12) = 0

(x-2)(x-6) ≥ 0

x-2 = 0

x = 2

x-6 = 0

x = 6

Now I will plug these values back into the inequality to see if x needs to be ≥ 2,6 or ≤ 2,6

In order for (x²–8x+12) ≥ 0 to be true, x ≥ 6, and x ≤ 2.

Therefore, the range of values that make the original inequality true is ( -∞ , 2 ] U [ 6 , ∞ )