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

1 Answer
Apr 1, 2018

Answer:

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!