How do you solve #4w^2+100=0# using the quadratic formula?

1 Answer
Apr 27, 2017

Answer:

There are no real solutions.
#w=-5i#
#w=5i#

Explanation:

The quadratic formula is as follows
#\frac{-b\pm\sqrt{b^2-4ac}}{2a}#
Given a function in the form
#aw^2+bw+c=0#

Plugging in the values we have, we the following
#\frac{-(0)+\sqrt{(0)^2-4(4)(100)}}{2(4)}#
#\frac{-(0)-\sqrt{(0)^2-4(4)(100)}}{2(4)}#

Solving for the first one we get
#\frac{\sqrt{0-1600}}{8}=\frac{\sqrt{-1600}}{8}#

Solving for the second one we get
#\frac{-\sqrt{0-1600}}{8}=\frac{-\sqrt{-1600}}{8}#

There are no real solutions, but we can get the imaginary/complex solutions
#\frac{\sqrt{1600\cdot -1}}{8}=\frac{40i}{8}=5i#
#\frac{-\sqrt{1600\cdot -1}}{8}=\frac{-40i}{8}=-5i#