# Fixing domains with mods

1. Jun 20, 2012

### Kartik.

mod(x-6) > mod(x^2 - 5x + 9)

Can anyone tell me about domain fixtures with mods using the above inequation?

A real beginner, so have mercy, be elementry .:D

2. Jun 20, 2012

### Curious3141

The word "mods" can have different meanings in Maths. I think you're using it in the sense of "modulus", or absolute value here.

So just to be clear, are you asking how to solve this inequality?

$$|x-6| > |x^2 - 5x + 9|$$

You should start by making a careful sketch of both curves. Remember to consider the discriminant of the quadratic expression, its minimum point and the fact that $|x-6|$ is actually represented by two lines at right angles to each other.

3. Jun 20, 2012

### HallsofIvy

I am going to assume you are using "mod" to mean "absolute value" (unfortunately "mod" has many different meanings in different circumstances.)

To find values of x for which $|x- 6|> |x^2- 5x+ 9|$ consider the possible signs for the quantities inside the absolute values. Solve x- 6> 0 and x^2- 5x+ 9> 0 to determine where those are positive or negative.

Where x- 6 and $x^2- 5x+ 9$ are both positive, the absolute values can be taken off to give $x- 6> x^2- 5x+ 9$ or $0> x^2- 5x+ 9$. Solve that inequality and take the values of x that satisfy it as well as $x- 6> 0$ and $x^2- 5x+ 9> 0$.

Where x- 6 is positive and $x^2- 5x+ 9$ is negative, the absolute values can be taken off to give $x- 6> -(x^2- 6x+ 9)$ which is the same as $x^2- 5x+ 3> 0$.

Where x- 6 is negative and $x^2- 5x+ 9$ is positive, the absolute values can be taken off to give $-(x- 6)> x^2- 6x+ 9$ which is the same as $0>x^2- 5x+ 3> 0$.

Where both x- 6 and $x^2- 5x+ 9$ are negative, the absolute values can be taken off to give $-(x- 6)> -(x^2- 6x+ 9)$ which is the same as $x^2- 7x+ 15> 0$.