How Do You Solve the Inequality |x-6| > |x^2-5x+9|?

[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

 

[Curious3141]

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

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.

 

[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.