How do you solve #x^2-3x-4<0#?

1 Answer
Jan 8, 2017

Answer:

Open interval (-1, 2)

Explanation:

#f(x) = x^2 - 3x - 4 < 0#
First, solve the quadratic equation f(x) = 0 to get the 2 real roots.
Since a - b + c = 0, use shortcut.
The 2 real roots are: x1 = - 1 and x= - c/a = 4/2 = 2
Since a = 1 > 0, the parabola graph open upward. Between the 2 real roots, f(x) < 0 as the graph stays below the x-axis.
Answer by open interval: (-1, 2). The 2 end points (critical points) aren't included in the solution set.
Graph on the number line:

------------------ - 1 ===== 0 ============ 2 ---------------------------