Accedmychevron_right11thchevron_rightmathchevron_rightTheory Of Quadratic Functionschevron_rightExercise 3.2

Solution

Given:

h(t)=5t2+20t+30h(t)=-5t^2+20t+30

We want h(t)40h(t)\ge 40:

5t2+20t+3040-5t^2+20t+30\ge 40 5t2+20t100-5t^2+20t-10\ge 0

Divide by 5-5 (reverse inequality):

t24t+20t^2-4t+2\le 0

Solve t24t+2=0t^2-4t+2=0:

t=4±1682=4±82=4±222=2±2\begin{aligned} t&=\frac{4\pm\sqrt{16-8}}{2}\\ &=\frac{4\pm\sqrt{8}}{2}\\ &=\frac{4\pm 2\sqrt{2}}{2}\\ &=2\pm\sqrt{2} \end{aligned}

Since the parabola opens upward, 0\le 0 holds between the roots:

22t2+2\boxed{2-\sqrt{2}\le t\le 2+\sqrt{2}} 0.586t3.414 seconds\boxed{0.586\lesssim t\lesssim 3.414\ \text{seconds}}