Question

if discrimant is negative, do u have 2 imginary solutions?

Answer 1

If the discriminant b2 – 4ac is negative, then there are no real solutions of the equation . (You need complex numbers to deal with this case properly.

Answer 2

if the discriment is
positive: two real solutions
zero: One repeating real solution
negative: no real solutions, two complex solutions

Answer 3

hmm answer choices only have two real solutions, no solutions whatever, and two negative solutions, which one u think best fits?

Answer 4

@leon549

Answer 5

no solution or two complex solution

Answer 6

the answer says no solution whatsoever, so i'm not sure which one to pick..

Answer 7

The problem is that both are technically inaccurate choices!
The quadratic equation doesn't have solutions in the \(real\) numbers, but the quadratic equation does have solutions in the \(complex\) numbers.
The solutions to the quadratic equation in the complex numbers are complex numbers and not just imaginary.

Note that imaginary numbers are numbers of the form \(ni\) (n multiplied by the imaginary unit \(i = \sqrt{-1}\). Complex numbers are of the form \(a+bi\).

When we look at the discriminant being less than zero (\(b^2 -4ac < 0\), then this links to the quadratic equation:
\(\displaystyle x = \frac{-b \pm \sqrt{\color{red}{b^2 -4ac}}}{2a} \)

The discriminant is the item underneath the square root. It is the behavior of the square root that tells us how the solutions behave, because a positive number under the square root gives a positive real number; a zero under the square root gives a zero, and the negative under the square root gives an imaginary number.

But see that if we split the fraction up, this is two complex numbers:
\( \displaystyle x = \frac{-b}{2a} \pm \frac{\sqrt{b^2 -4ac}}{2a} = A + B \ i\)

To summarize, I think 'two imaginary numbers' is the \(more\) correct solution, but neither are necessarily correct.