Question

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

Answer 1

[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 ---------------------------