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

May 28, 2018

$9 {x}^{2} - 6 x + 1 = 9 {\left(x - \frac{1}{3}\right)}^{2} \le 0$ when $x = \frac{1}{3}$.

#### Explanation:

Actually the left side can never be less than 0 for real numbers. It's lowest value is $f \left(x\right) = 0$ for $x = \frac{1}{3}$

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 $x = \frac{1}{3}$ has the inclination $0$, i.e. is horisontal. Therefore the lowest point of the left side is here.

Other than that we can write:
$9 {x}^{2} - 6 x + 1 = 9 \left({x}^{2} - \frac{2}{3} x + {\left(\frac{1}{3}\right)}^{2}\right) = 9 {\left(x - \frac{1}{3}\right)}^{2}$
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 $x = \frac{1}{3}$.