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

##### 2 Answers

#### Answer:

#### Explanation:

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 **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, *positive* or both *negative*.

For any value of

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

For any value of

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

This means that your solution set will be **are not** valid solutions for this inequality because they will make the left-hand side product equal to zero.

#### Answer:

Solve quadratic inequality: f(x) = x^2 - 3x + 2 > 0

Ans: (-infinity, 1) and (2, +infinity)

#### Explanation:

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]}