Question

How do you find the roots, real and imaginary, of `h=-16t^2+64t` using the quadratic formula?

Answer 1

0, 4

Explanation:

The quadratic formula states
`x=(-b±√b^2-4ac)/(2a)`

In this case,
`a=-16`
`b=64`
`c=0`

Plug in the numbers.

To make it simpler, let's only substitute `a` first.

`x=(-b±√b^2-(-64c))/-32`

Substitute `b`.

`x=(-64±√64^2+64c)/-32`

Finally, substitute `c`.

`x=(-64±√64^2)/-32`

Simplify to get
`x=(-64±64)/-32`

If `x=(-64+64)/-32`, then `x=0` because `0/-32 = 0`.

If `x=(-64-64)/-32`, then `x=4` because `-128/-32=4`.

The roots of `h=-16t^2+64t` are 0, 4.

Yay!