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

Question

1971 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-01T21:55:35

0, 4

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!

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