Question

How do you solve the inequality `9x^2-6x+1<=0`?

Answer 1

`9x^2-6x+1=9(x-1/3)^2<=0` when `x=1/3`.

Explanation:

Actually the left side can never be less than 0 for real numbers. It's lowest value is `f(x)=0` for `x=1/3`

You can see that from a diagram:

Since this is precalculus, I'm in doubt if derivation should be used in the solution, but using it you can show that a tangent at `x=1/3` has the inclination `0`, i.e. is horisontal. Therefore the lowest point of the left side is here.

Other than that we can write:
`9x^2-6x+1=9(x^2-2/3x+(1/3)^2)=9(x-1/3)^2`
Since the left hand of the inequality is a square, we can conclude that it will never be negative, and it's lowest value is `0` when `x=1/3`.