How do you solve the inequality: #x^2 - 3x + 2 > 0#?

You can make take the quadratic and make it equal to zero in order to find its roots. This will help you determine the intervals for which it will be greater than zero.

So, for

#x^2 - 3x + 2 = 0#

use the quadratic formula to find

#x_(1,2) = (-(-3) +- sqrt((3-^2 - 4 * 1 * 2)))/(2 * 1)#

#x_(1,2) = (3 +- sqrt(1))/2#

#x_(1,2) = (3 +- 1)/2 = {(x_1 = (3 + 1)/2 = 2), (x_2 = (3-1)/2 = 1) :}#

For a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

you can rewrite it using its roots by using the formula

#color(blue)(a(x-x_1)(x-x_2) = 0)#

In your case, you have

#(x-2)(x-1) = 0#

Now you need to find the values of #x# that will make this greater than zero

#(x-2)(x-1) > 0#

In order for the left-hand side of this inequality to be positive, you need both terms, #(x-2)# and #(x-1)# to either be both positive or both negative.

For any value of #x>2# you will get

#{(x-2>0), (x-1>0):} implies (x-2)(x-1)>0#

For any value of #x<1# you will get

#{(x - 2<0), (x-1<0) :} implies (x-2)(x-1)>0#

This means that your solution set will be #x in (-oo, 1) uu (2, +oo)#. Remember that #x=1# and #x=2# are not valid solutions for this inequality because they will make the left-hand side product equal to zero.

First solve f(x) = x^2 - 3x + 2 = 0
Since (a + b + c = 0), use the Shortcut. The 2 real roots are x = 1 and x = c/a = 2.
Use the algebraic method to solve f(x) > 0. Since a > 0, the parabola opens upward. Inside the interval (1, 2), f(x) is negative. Outside the interval (1, 2), f(x) is positive.
Answer by open intervals: (-infinity, 1) and (2, +infinity)
graph{x^2 - 3x + 2 [-10, 10, -5, 5]}