How can I make a quadratic formula in standard form that has two complex solutions.

Question

anonymous · Answer

x = [-b +- sqrt(b^2 - 4ac]/2a two roots, root1 and root2

choose 4 a c > b^2

roots come from (x-root1)(x-root2) = f(x)

anonymous · Answer

I'm sorry but this makes no sense to me. Can you explain it in a simpler way? please?!

anonymous · Answer

@douglaswinslowcooper

anonymous · Answer

quadratic formula is a x^2 + bx + c = 0
standard form.

it can be factored into (x-root1)(x-root2) = 0

the formula I gave you gives the two roots (note + or - sign)

Your problem is that to make these have imaginary roots, you need b^2-4ac<0.

As you are not asked for roots, just pick a, b, and c to satisfy this inequality.

anonymous · Answer

Glad you got it.

anonymous · Answer

@douglaswinslowcooper  thanks a lot!!