How do you solve the inequality: #x^2 - 3x + 2 > 0#?
2 Answers
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
#(x-2)(x-1) > 0#
In order for the left-hand side of this inequality to be positive, you need both terms,
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
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]}