Question

1. $x^{2}+x - 8 = 0$ $1^{2}-4(1)(-8)=33$ 2 real solutions 6. $3x^{2}+36 = 0$ $0^{2}-4(3)(36)=-432$ 2 imaginary solution 7. $3x^{2}-6x + 11 = 0$ $-6^{2}-4(3)(11)=-96$ 2 imaginary solutions 8. $4x^{2}-6x=-2$ 9. a quarterback throws a football to a receiver. the path of a football can be modeled by the quadratic function $h=-16t^{2}+45t + 4$, where $h$ is the height in feet and $t$ is the number of seconds after the football is thrown. if the ball is overthrown and the receiver does not touch the ball, how long will it take the football to hit the ground? 10. the height $h(t)$ in feet of an object $t$ seconds after it is propelled up from the ground with an initial velocity of 60 feet per second is modeled by the equation $h(t)=-16t^{2}+60t$. when will the object be at a height of 56 feet?

Explanation:

Step1: Set up the equation

When the football hits the ground, \(h = 0\). So we set \(-16t^{2}+45t + 4=0\).

Step2: Identify coefficients

For the quadratic equation \(ax^{2}+bx + c = 0\), here \(a=-16\), \(b = 45\), \(c = 4\).

Step3: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute the values: \(b^{2}-4ac=(45)^{2}-4\times(-16)\times4=2025 + 256=2281\).
Then \(t=\frac{-45\pm\sqrt{2281}}{2\times(-16)}=\frac{-45\pm47.76}{-32}\).

Step4: Calculate the two values of \(t\)

\(t_1=\frac{-45 + 47.76}{-32}=\frac{2.76}{-32}\approx - 0.086\) (discarded as time cannot be negative).
\(t_2=\frac{-45-47.76}{-32}=\frac{-92.76}{-32}\approx2.9\).

Answer:

It will take approximately \(2.9\) seconds for the football to hit the ground.