Question

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

Answer 1

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`