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

Question

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

Nathan Axel · Accepted Answer

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

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

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]}

Hazel Anglin · Answer

#x in (-oo, 1) uu (2, + oo)#

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.

Eva Abramson · Answer

undefined To solve the inequality x^2 - 3x + 2 > 0, first factor the quadratic expression:

x^2 - 3x + 2 = (x - 2)(x - 1)

Next, find the critical points by setting each factor equal to zero:

x - 2 = 0 → x = 2
x - 1 = 0 → x = 1

Now, create intervals on the number line using these critical points:

Interval 1: (-∞, 1)
Interval 2: (1, 2)
Interval 3: (2, ∞)

Now, test a value from each interval in the original inequality:

For Interval 1, choose x = 0:
0^2 - 3(0) + 2 = 2 > 0

For Interval 2, choose x = 1.5:
(1.5)^2 - 3(1.5) + 2 = 0.25 - 4.5 + 2 = -2.25 < 0

For Interval 3, choose x = 3:
3^2 - 3(3) + 2 = 9 - 9 + 2 = 2 > 0

Since the inequality is greater than zero for Interval 1 and Interval 3, but less than zero for Interval 2, the solution to the inequality is:

x ∈ (-∞, 1) U (2, ∞)

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

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