How do you solve the inequality #9x^2-6x+1

Question

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

Serenity Johnson · Accepted Answer

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

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#.

James Allen · Answer

undefined To solve the inequality $9x^2 - 6x + 1 \leq 0$, you can follow these steps:

1. Find the roots of the quadratic equation $9x^2 - 6x + 1 = 0$. You can use the quadratic formula or factoring.
2. Once you have the roots, use them to determine the intervals where the quadratic expression is less than or equal to zero.
3. Test a value within each interval to determine whether the expression is positive or negative.
4. Identify the intervals where the expression is less than or equal to zero.

Then, you'll have the solution to the inequality in terms of intervals on the number line.

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

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