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

Aug 3, 2016

[p1 , p2]

#### Explanation:

Bring the quadratic inequality to standard form:
$f \left(p\right) = - 5 {p}^{2} + 4 p + 4 \ge 0$.
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):
$D = {d}^{2} = 16 + 64 = 80 = 5 \left(16\right)$ --> $d = \pm 4 \sqrt{5}$
There are 2 x-intercepts:
$p = - \frac{4}{-} 10 \pm 4 \frac{\sqrt{5}}{10} = \frac{2}{5} \pm \frac{2 \sqrt{5}}{5} = \frac{2 \pm \sqrt{5}}{5}$
$p 1 = \frac{2 - \sqrt{5}}{5}$ and $p 2 = \frac{2 + \sqrt{5}}{5}$.
Answer by closed interval: [p1 , p2]. The 2 end points p1 and p2 are included in the solution set.
Graph:

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