How do you solve #-2x^2+6X+15 <=0#?

1 Answer
Jun 5, 2016

Answer:

Half closed intervals: (-inf., x2] and [x1, inf.)

Explanation:

#f(x) - 2x^2 + 6x + 15 <= 0#
First, solve f(x) = 0 to find the 2 real roots. Use the improved quadratic formula (Socratic Search)
#D = d^2 = b^2 - 4ac = 36 + 120 = 156# --> #d = 2sqrt39#.
There are 2 real roots (2 x-intercepts):
#x = -b/(2a) +- d/(2a) = 6/-4 +- sqrt39/2 = (3 +- sqrt39)/2#
#x1 = (3 + sqrt39)/2# and #x2 = (3 - sqrt39)/2#.
Since a < 0, the parabola opens downward.
f(x) > 0 inside the interval (x2, x1), between the 2 x-intercepts.
f(x) < 0 outside the (x2, x1) interval, because the 2 parabola branches are below the x-axis.
Solution set: Two half-closed intervals (-inf., x2] and [x1, inf.)
The 2 end points x1 and x2 are include in the solution set