How do you solve #4+4p-5p^2>=0#?

1 Answer
Aug 3, 2016

Answer:

[p1 , p2]

Explanation:

Bring the quadratic inequality to standard form:
#f(p) = -5p^2 + 4p + 4 >= 0#.
Since a< 0, the parabola opens downward. Between the 2 x-intercepts, f(p) >= 0, as a part of the parabola stays over the x-axis.
Find the 2 x-intercepts (real roots) by the new quadratic formula in graphic form (Socratic Search):
#D = d^2 = 16 + 64 = 80= 5(16)# --> #d = +- 4sqrt5#
There are 2 x-intercepts:
#p = -4/-10 +- 4sqrt5/10 = 2/5 +- (2sqrt5)/5 = (2 +- sqrt5)/5#
#p1 = (2 - sqrt5)/5# and #p2 = (2 + sqrt5)/5#.
Answer by closed interval: [p1 , p2]. The 2 end points p1 and p2 are included in the solution set.
Graph:

------------------------p1 === 0 ============ p2 ---------------------------