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

1 Answer
May 28, 2018

Answer:

#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:
enter image source here

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#.